Ex 5.8 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution: Differentiating both sides of the equation \[ x – y = \pi \] with respect to \(x\), we get \[ \frac{d}{dx}(x) – \frac{d}{dx}(y) = \frac{d}{dx}(\pi). \] Since \(\frac{d}{dx}(x) = 1\), \(\frac{d}{dx}(y) = \frac{dy}{dx}\) and the derivative of a constant is zero, this becomes \[ 1 – \frac{dy}{dx} = 0. \] Solving for \(\frac{dy}{dx}\), \[ \frac{dy}{dx} = 1. \]

Solution: Differentiating both sides of \[ x^2 + y^2 = 25 \] with respect to \(x\), we obtain \[ 2x + 2y\frac{dy}{dx} = 0. \] Solving for \(\frac{dy}{dx}\), we have \[ 2y\frac{dy}{dx} = -2x \quad \Longrightarrow \quad \frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}. \]

Solution: Differentiating both sides of \[ xy = c^2 \] with respect to \(x\), we use the product rule on the left-hand side: \[ \frac{d}{dx}(xy) = y + x\frac{dy}{dx}. \] Since \(c^2\) is a constant, its derivative is zero: \[ y + x\frac{dy}{dx} = 0. \] Solving for \(\frac{dy}{dx}\), we subtract \(y\) from both sides: \[ x\frac{dy}{dx} = -y, \] and then divide both sides by \(x\): \[ \frac{dy}{dx} = -\frac{y}{x}. \]

Solution: Differentiating both sides of \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2016 \] with respect to \(x\), we get: \[ \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0. \] Isolating \(\frac{dy}{dx}\), we have: \[ \frac{2y}{b^2}\frac{dy}{dx} = -\frac{2x}{a^2}, \] and dividing both sides by \(\frac{2y}{b^2}\) yields: \[ \frac{dy}{dx} = -\frac{2x}{a^2} \cdot \frac{b^2}{2y} = -\frac{b^2 x}{a^2 y}. \]

Solution: Differentiating both sides of \[ \sqrt{x} + \sqrt{y} = \sqrt{c} \] with respect to \(x\) gives \[ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0. \] Multiplying through by \(2\) to simplify, we have \[ \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} \frac{dy}{dx} = 0. \] Isolating \(\frac{dy}{dx}\), we obtain \[ \frac{1}{\sqrt{y}} \frac{dy}{dx} = -\frac{1}{\sqrt{x}}, \] and then multiplying both sides by \(\sqrt{y}\) yields \[ \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}. \]

Solution: Differentiate both sides of \[ x^3 + x^2y + xy^2 + y^3 = 81 \] with respect to \(x\). Differentiating term by term: \[ \begin{aligned} \frac{d}{dx}(x^3) &= 3x^2, \\ \frac{d}{dx}(x^2y) &= 2xy + x^2\frac{dy}{dx}, \quad \text{(using the product rule)}\\ \frac{d}{dx}(xy^2) &= y^2 + 2xy\frac{dy}{dx}, \quad \text{(using the product rule)}\\ \frac{d}{dx}(y^3) &= 3y^2\frac{dy}{dx}. \end{aligned} \] Combining these, we have: \[ 3x^2 + 2xy + x^2\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0. \] Grouping the terms containing \(\frac{dy}{dx}\) together: \[ 3x^2 + 2xy + y^2 + \left(x^2 + 2xy + 3y^2\right)\frac{dy}{dx} = 0. \] Isolating \(\frac{dy}{dx}\), we get: \[ \left(x^2 + 2xy + 3y^2\right)\frac{dy}{dx} = -\left(3x^2 + 2xy + y^2\right), \] and hence, \[ \frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}. \]

Solution: Differentiate both sides of \[ 2x+3y=\sin x \] with respect to \(x\). The derivative of the left-hand side is: \[ \frac{d}{dx}(2x)+\frac{d}{dx}(3y)=2+3\frac{dy}{dx}, \] and the derivative of the right-hand side is: \[ \frac{d}{dx}(\sin x)=\cos x. \] Setting these equal, we have: \[ 2+3\frac{dy}{dx}=\cos x. \] Solving for \(\frac{dy}{dx}\), subtract \(2\) from both sides: \[ 3\frac{dy}{dx}=\cos x-2, \] and then divide both sides by \(3\): \[ \frac{dy}{dx}=\frac{\cos x-2}{3}. \]

Solution: Differentiate both sides of the equation with respect to \( x \): \[ \frac{d}{dx}(2x+3y)= \frac{d}{dx}(\sin y). \] On the left-hand side: \[ \frac{d}{dx}(2x)=2 \quad \text{and} \quad \frac{d}{dx}(3y)=3\frac{dy}{dx}, \] so it becomes: \[ 2+3\frac{dy}{dx}. \] On the right-hand side, applying the chain rule: \[ \frac{d}{dx}(\sin y)=\cos y\cdot\frac{dy}{dx}. \] Therefore, the differentiated equation is: \[ 2+3\frac{dy}{dx}=\cos y\cdot\frac{dy}{dx}. \] Rearranging the terms: \[ 2=\frac{dy}{dx}(\cos y-3). \] Solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx}=\frac{2}{\cos y-3}. \]

Solution: Differentiate both sides with respect to \( x \). Let \[ u=x^2+y^2. \] Then the left-hand side becomes \( u^2 \) whose derivative is \[ \frac{d}{dx}(u^2)=2u\frac{du}{dx}. \] Now, since \[ \frac{du}{dx}=\frac{d}{dx}(x^2+y^2)=2x+2y\frac{dy}{dx}, \] we have: \[ \frac{d}{dx}\left[\left(x^2+y^2\right)^2\right] = 2(x^2+y^2)(2x+2y\frac{dy}{dx}) = 4(x^2+y^2)(x+y\frac{dy}{dx}). \] The right-hand side is the product \(xy\). Differentiate using the product rule: \[ \frac{d}{dx}(xy)=y+x\frac{dy}{dx}. \] Thus, the differentiated equation is: \[ 4(x^2+y^2)(x+y\frac{dy}{dx})=y+x\frac{dy}{dx}. \] Now, expand and collect the terms containing \( \frac{dy}{dx} \): \[ 4x(x^2+y^2)+4y(x^2+y^2)\frac{dy}{dx}=y+x\frac{dy}{dx}. \] Rearranging gives: \[ 4y(x^2+y^2)\frac{dy}{dx} – x\frac{dy}{dx} = y-4x(x^2+y^2). \] Factor \( \frac{dy}{dx} \): \[ \frac{dy}{dx}\left[4y(x^2+y^2)-x\right]=y-4x(x^2+y^2). \] Finally, solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx}=\frac{y-4x(x^2+y^2)}{4y(x^2+y^2)-x}. \]

Solution: Differentiate both sides of the equation with respect to \( x \). The left-hand side is \( \sin (x+y) \). Using the chain rule: \[ \frac{d}{dx}[\sin (x+y)] = \cos (x+y)\cdot\frac{d}{dx}(x+y) = \cos (x+y)\cdot\left(1+\frac{dy}{dx}\right). \] The right-hand side is a constant: \[ \frac{d}{dx}\left(\frac{2}{3}\right)=0. \] Therefore, the differentiated equation is: \[ \cos (x+y)\cdot\left(1+\frac{dy}{dx}\right)=0. \] Since \( \cos (x+y) \neq 0 \) (as \( \sin (x+y)=\frac{2}{3} \) ensures \( x+y \) is such that the cosine is non-zero), we can divide by \( \cos (x+y) \): \[ 1+\frac{dy}{dx}=0. \] Hence, solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx}=-1. \]

Solution: Differentiate the equation \( x^{2/3}+y^{2/3}=2 \) implicitly with respect to \( x \): \[ \frac{d}{dx}\left(x^{2/3}\right) + \frac{d}{dx}\left(y^{2/3}\right)=\frac{d}{dx}(2). \] The derivative of \( x^{2/3} \) is: \[ \frac{2}{3}x^{-1/3}. \] For \( y^{2/3} \), applying the chain rule: \[ \frac{d}{dx}\left(y^{2/3}\right)=\frac{2}{3}y^{-1/3}\frac{dy}{dx}. \] The right-hand side derivative is 0. Hence, we have: \[ \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx}=0. \] Divide the entire equation by \(\frac{2}{3}\): \[ x^{-1/3}+y^{-1/3}\frac{dy}{dx}=0. \] Rearranging for \(\frac{dy}{dx}\): \[ y^{-1/3}\frac{dy}{dx}=-x^{-1/3} \quad \Longrightarrow \quad \frac{dy}{dx}=-\frac{x^{-1/3}}{y^{-1/3}}. \] Simplify the expression by writing: \[ \frac{dy}{dx}=-\frac{y^{1/3}}{x^{1/3}}. \] Now, substitute the point \((1,1)\): \[ \frac{dy}{dx}=-\frac{1^{1/3}}{1^{1/3}}=-1. \]

Solution: First, differentiate \( y^2=4ax \) with respect to \( x \): \[ 2y\frac{dy}{dx}=4a \quad \Longrightarrow \quad \frac{dy}{dx}=\frac{4a}{2y}=\frac{2a}{y}. \] Next, differentiate the same equation with respect to \( y \), treating \( x \) as a function of \( y \): \[ \frac{d}{dy}(y^2)=\frac{d}{dy}(4ax) \quad \Longrightarrow \quad 2y = 4a\frac{dx}{dy}. \] Solve for \( \frac{dx}{dy} \): \[ \frac{dx}{dy}=\frac{2y}{4a}=\frac{y}{2a}. \] Now, multiply \( \frac{dy}{dx} \) and \( \frac{dx}{dy} \): \[ \frac{dy}{dx}\cdot\frac{dx}{dy}=\frac{2a}{y}\cdot\frac{y}{2a}=1. \]

Solution: We start with the equation: \[ x^3 + y^3 = 3axy. \] Differentiate both sides with respect to \( x \): Left-hand side: \[ \frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = 3x^2 + 3y^2\frac{dy}{dx}. \] Right-hand side (product rule on \( 3axy \)): \[ \frac{d}{dx}(3axy) = 3a\left(y + x\frac{dy}{dx}\right). \] So the differentiated equation becomes: \[ 3x^2 + 3y^2\frac{dy}{dx} = 3a\left(y + x\frac{dy}{dx}\right). \] Expand the right-hand side: \[ 3x^2 + 3y^2\frac{dy}{dx} = 3ay + 3ax\frac{dy}{dx}. \] Move all terms to one side: \[ 3y^2\frac{dy}{dx} – 3ax\frac{dy}{dx} = 3ay – 3x^2. \] Factor out \( \frac{dy}{dx} \): \[ (3y^2 – 3ax)\frac{dy}{dx} = 3ay – 3x^2. \] Divide both sides: \[ \frac{dy}{dx} = \frac{3ay – 3x^2}{3y^2 – 3ax}. \] Simplify: \[ \frac{dy}{dx} = \frac{ay – x^2}{y^2 – ax}. \] Now, take reciprocal to get \( \frac{dx}{dy} \): \[ \frac{dx}{dy} = \frac{y^2 – ax}{ay – x^2}. \] Multiply: \[ \frac{dy}{dx} \cdot \frac{dx}{dy} = \left( \frac{ay – x^2}{y^2 – ax} \right) \cdot \left( \frac{y^2 – ax}{ay – x^2} \right) = 1. \]

Solution: Differentiate the given equation with respect to \( x \): \[ \frac{d}{dx}\left(y+\sin y\right)=\frac{d}{dx}\left(\cos x\right). \] The left-hand side becomes: \[ \frac{dy}{dx}+\cos y\frac{dy}{dx}=\left(1+\cos y\right)\frac{dy}{dx}, \] while the right-hand side is: \[ -\sin x. \] Hence, we have: \[ \left(1+\cos y\right)\frac{dy}{dx}=-\sin x. \] Solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx}=-\frac{\sin x}{1+\cos y}. \]

Solution: Differentiate both sides of the equation with respect to \( x \). For the left-hand side: \[ \frac{d}{dx}(ax)=a \quad \text{and} \quad \frac{d}{dx}(by^2)=2by\frac{dy}{dx}. \] For the right-hand side, applying the chain rule: \[ \frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}. \] Therefore, the differentiated equation becomes: \[ a+2by\frac{dy}{dx}=-\sin y\frac{dy}{dx}. \] Rearranging to collect terms involving \( \frac{dy}{dx} \): \[ 2by\frac{dy}{dx}+\sin y\frac{dy}{dx}=-a. \] Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx}(2by+\sin y)=-a. \] Solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx}=-\frac{a}{2by+\sin y}. \]

Solution: We differentiate the equation implicitly with respect to \( x \). The given equation is: \[ y\sec x + \tan x + x^2y = 0. \] \(\textbf{Step 1: Differentiate each term}\) \(\bullet\) For \( y\sec x \), using the product rule: \[ \frac{d}{dx}(y\sec x)=\sec x\frac{dy}{dx}+y\sec x\tan x. \] \(\bullet\) For \( \tan x \): \[ \frac{d}{dx}(\tan x)=\sec^2 x. \] \(\bullet\) For \( x^2y \), again using the product rule: \[ \frac{d}{dx}(x^2y)=2xy+x^2\frac{dy}{dx}. \] \(\textbf{Step 2: Write the differentiated equation}\) Combining the derivatives, we have: \[ \sec x\frac{dy}{dx}+y\sec x\tan x + \sec^2 x + 2xy + x^2\frac{dy}{dx}=0. \] \(\textbf{Step 3: Group the terms containing } \frac{dy}{dx}\) Factor \( \frac{dy}{dx} \): \[ \left(\sec x+x^2\right)\frac{dy}{dx} + \left(y\sec x\tan x+\sec^2 x+2xy\right)=0. \] \(\textbf{Step 4: Solve for } \frac{dy}{dx}\) Isolate \( \frac{dy}{dx} \): \[ \left(\sec x+x^2\right)\frac{dy}{dx} = -\left(y\sec x\tan x+\sec^2 x+2xy\right), \] and thus, \[ \frac{dy}{dx}=-\frac{y\sec x\tan x+\sec^2 x+2xy}{\sec x+x^2}. \]

Solution: Differentiate both sides of the equation with respect to \( x \): \[ \frac{d}{dx}\left(\sin^2 x+\cos^2 y\right)=\frac{d}{dx}(1). \] The right-hand side is 0. \(\textbf{Step 1: Differentiate } \sin^2 x\) Using the chain rule: \[ \frac{d}{dx}\left(\sin^2 x\right)=2\sin x\cos x. \] \(\textbf{Step 2: Differentiate } \cos^2 y\) Again, using the chain rule: \[ \frac{d}{dx}\left(\cos^2 y\right)=2\cos y\cdot(-\sin y)\cdot\frac{dy}{dx}=-2\cos y\sin y\frac{dy}{dx}. \] \(\textbf{Step 3: Form the Equation}\) The differentiated equation becomes: \[ 2\sin x\cos x-2\cos y\sin y\frac{dy}{dx}=0. \] \(\textbf{Step 4: Solve for } \frac{dy}{dx}\) Rearranging: \[ 2\cos y\sin y\frac{dy}{dx}=2\sin x\cos x, \] so that: \[ \frac{dy}{dx}=\frac{2\sin x\cos x}{2\sin y\cos y}=\frac{\sin x\cos x}{\sin y\cos y}. \] Recognizing the double-angle identities: \[ \sin 2x=2\sin x\cos x \quad \text{and} \quad \sin 2y=2\sin y\cos y, \] we can write: \[ \frac{dy}{dx}=\frac{\sin 2x}{\sin 2y}. \]

Solution: We start with the given equation: \[ xy + y^{2} = \tan x + y. \] Differentiating both sides with respect to \( x \): \[ \frac{d}{dx}(xy) + \frac{d}{dx}(y^{2}) = \frac{d}{dx}(\tan x) + \frac{d}{dx}(y). \] Note that: \[ \frac{d}{dx}(xy) = y + x\frac{dy}{dx}, \quad \frac{d}{dx}(y^{2}) = 2y\frac{dy}{dx}, \] \[ \frac{d}{dx}(\tan x) = \sec^{2} x, \quad \frac{d}{dx}(y) = \frac{dy}{dx}. \] Substituting these derivatives into the equation, we obtain: \[ y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = \sec^{2} x + \frac{dy}{dx}. \] Combine the terms containing \(\frac{dy}{dx}\): \[ \left( x + 2y – 1 \right)\frac{dy}{dx} = \sec^{2} x – y. \] Finally, solving for \(\frac{dy}{dx}\) gives: \[ \frac{dy}{dx} = \frac{\sec^{2} x – y}{x + 2y – 1}. \]

Solution: Differentiating both sides of the equation with respect to \( x \), we get: \[ \frac{d}{dx}(y) = \frac{d}{dx}\left(\tan(x+y)\right). \] The left side is: \[ \frac{dy}{dx}. \] The right side, using the chain rule, is: \[ \sec^{2}(x+y) \cdot \left(1+\frac{dy}{dx}\right). \] Thus, we have: \[ \frac{dy}{dx} = \sec^{2}(x+y)\left(1+\frac{dy}{dx}\right). \] Rearranging to solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} – \sec^{2}(x+y)\frac{dy}{dx} = \sec^{2}(x+y), \] \[ \frac{dy}{dx}\left(1-\sec^{2}(x+y)\right) = \sec^{2}(x+y), \] \[ \frac{dy}{dx} = \frac{\sec^{2}(x+y)}{1-\sec^{2}(x+y)}. \] Noting that \(1-\sec^{2}(x+y) = -\tan^{2}(x+y)\), we obtain: \[ \frac{dy}{dx} = -\frac{\sec^{2}(x+y)}{\tan^{2}(x+y)}. \] Since \[ \frac{\sec^{2}(x+y)}{\tan^{2}(x+y)} = \frac{1/\cos^{2}(x+y)}{\sin^{2}(x+y)/\cos^{2}(x+y)} = \frac{1}{\sin^{2}(x+y)} = \text{cosec}^{2}(x+y), \] we have: \[ \frac{dy}{dx} = -\text{cosec}^{2}(x+y). \]

Solution: Starting with the given equation: \[ \sin(xy)+\frac{x}{y}= x^{2}-y. \] Differentiate both sides with respect to \( x \). For the left-hand side, we differentiate term by term: \[ \frac{d}{dx}\left(\sin(xy)\right) + \frac{d}{dx}\left(\frac{x}{y}\right). \] For \(\sin(xy)\), using the chain rule: \[ \frac{d}{dx}\left(\sin(xy)\right) = \cos(xy)\cdot\frac{d}{dx}(xy) = \cos(xy)\left(y+x\frac{dy}{dx}\right). \] For \(\frac{x}{y}\), rewriting as \(x y^{-1}\) and differentiating: \[ \frac{d}{dx}\left(xy^{-1}\right)= \frac{1}{y} – \frac{x}{y^{2}}\frac{dy}{dx}. \] The derivative of the right-hand side \(x^2-y\) is: \[ \frac{d}{dx}(x^2-y)= 2x-\frac{dy}{dx}. \] Combining these, we have: \[ \cos(xy)\left(y+x\frac{dy}{dx}\right) + \frac{1}{y}-\frac{x}{y^{2}}\frac{dy}{dx} = 2x-\frac{dy}{dx}. \] Grouping the terms containing \(\frac{dy}{dx}\) on one side: \[ \cos(xy)y + \frac{1}{y} + \left(\cos(xy)x-\frac{x}{y^{2}}\right)\frac{dy}{dx} = 2x-\frac{dy}{dx}. \] Bring the \(\frac{dy}{dx}\) term from the right-hand side to the left: \[ \cos(xy)y + \frac{1}{y} + \left(\cos(xy)x-\frac{x}{y^{2}}+\;1\right)\frac{dy}{dx} = 2x. \] Finally, solving for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{2x-\cos(xy)y-\frac{1}{y}}{\cos(xy)x-\frac{x}{y^{2}}+1}. \]

Solution: We start with the equation: \[ \sec(x+y)=xy. \] Differentiating both sides with respect to \( x \): \[ \frac{d}{dx}\left(\sec(x+y)\right)=\frac{d}{dx}(xy). \] Using the chain rule on the left-hand side: \[ \frac{d}{dx}\left(\sec(x+y)\right)=\sec(x+y)\tan(x+y)\left(1+\frac{dy}{dx}\right). \] The derivative of the right-hand side is: \[ \frac{d}{dx}(xy)= y+x\frac{dy}{dx}. \] Thus, we have: \[ \sec(x+y)\tan(x+y)\left(1+\frac{dy}{dx}\right)= y+x\frac{dy}{dx}. \] Expanding the left-hand side: \[ \sec(x+y)\tan(x+y)+\sec(x+y)\tan(x+y)\frac{dy}{dx}= y+x\frac{dy}{dx}. \] Grouping the terms containing \(\frac{dy}{dx}\): \[ \sec(x+y)\tan(x+y)\frac{dy}{dx}-x\frac{dy}{dx}= y-\sec(x+y)\tan(x+y). \] Factor out \(\frac{dy}{dx}\): \[ \left(\sec(x+y)\tan(x+y)-x\right)\frac{dy}{dx}= y-\sec(x+y)\tan(x+y). \] Finally, solving for \(\frac{dy}{dx}\) gives: \[ \frac{dy}{dx}=\frac{y-\sec(x+y)\tan(x+y)}{\sec(x+y)\tan(x+y)-x}. \]

Solution: We start with the given equation: \[ \sec\left(\frac{x+y}{x-y}\right)=a. \] Since \(a\) is a constant, its derivative with respect to \(x\) is zero: \[ \frac{d}{dx}\left[\sec\left(\frac{x+y}{x-y}\right)\right] = 0. \] Using the chain rule, the derivative of the left-hand side is: \[ \sec\left(\frac{x+y}{x-y}\right)\tan\left(\frac{x+y}{x-y}\right) \cdot \frac{d}{dx}\left(\frac{x+y}{x-y}\right)=0. \] Since \(\sec\left(\frac{x+y}{x-y}\right)\tan\left(\frac{x+y}{x-y}\right)\) is generally non-zero, we must have: \[ \frac{d}{dx}\left(\frac{x+y}{x-y}\right)=0. \] Let \(u=\frac{x+y}{x-y}\). Differentiating using the quotient rule: \[ \frac{du}{dx}=\frac{(1+\frac{dy}{dx})(x-y)-(x+y)(1-\frac{dy}{dx})}{(x-y)^2}=0. \] Expanding the numerator: \[ (1+\frac{dy}{dx})(x-y) = x-y+x\frac{dy}{dx}-y\frac{dy}{dx}, \] \[ (1-\frac{dy}{dx})(x+y) = x+y-x\frac{dy}{dx}-y\frac{dy}{dx}. \] Therefore, the numerator becomes: \[ \Bigl[x-y+x\frac{dy}{dx}-y\frac{dy}{dx}\Bigr] – \Bigl[x+y-x\frac{dy}{dx}-y\frac{dy}{dx}\Bigr]=0. \] Simplifying: \[ (x-y – x-y) + \left(x\frac{dy}{dx} – y\frac{dy}{dx}+x\frac{dy}{dx}+y\frac{dy}{dx}\right)= -2y+2x\frac{dy}{dx}=0. \] This simplifies to: \[ 2x\frac{dy}{dx} = 2y. \] Dividing both sides by \(2x\) (assuming \(x\neq 0\)): \[ \frac{dy}{dx}=\frac{y}{x}. \]

Solution: We start with the equation: \[ y = x \sin (a+y). \] Differentiating both sides with respect to \( x \): \[ \frac{dy}{dx} = \sin(a+y) + x\cos(a+y) \cdot \frac{dy}{dx}. \] Rearranging to collect the \(\frac{dy}{dx}\) terms: \[ \frac{dy}{dx} – x\cos(a+y)\frac{dy}{dx} = \sin(a+y), \] \[ \frac{dy}{dx}\left(1-x\cos(a+y)\right) = \sin(a+y). \] Solving for \(\frac{dy}{dx}\) gives: \[ \frac{dy}{dx} = \frac{\sin(a+y)}{1-x\cos(a+y)}. \] From the original equation, we have: \[ y = x\sin(a+y) \quad \Longrightarrow \quad x = \frac{y}{\sin(a+y)}. \] Substituting this into the denominator: \[ 1-x\cos(a+y) = 1-\frac{y}{\sin(a+y)}\cos(a+y) = \frac{\sin(a+y)-y\cos(a+y)}{\sin(a+y)}. \] Therefore, \[ \frac{dy}{dx} = \frac{\sin(a+y)}{\frac{\sin(a+y)-y\cos(a+y)}{\sin(a+y)}} = \frac{\sin^2(a+y)}{\sin(a+y)-y\cos(a+y)}. \]

Solution: Starting with the given equation: \[ \sin x = y \sin(x+a), \] we differentiate both sides with respect to \( x \). Differentiating the left-hand side: \[ \frac{d}{dx}(\sin x)=\cos x. \] For the right-hand side, applying the product rule: \[ \frac{d}{dx}\bigl(y \sin(x+a)\bigr)=\frac{dy}{dx}\sin(x+a)+y\cos(x+a). \] Thus, we have: \[ \cos x=\frac{dy}{dx}\sin(x+a)+y\cos(x+a). \] From the original equation, we express \( y \) as: \[ y=\frac{\sin x}{\sin(x+a)}. \] Substituting this value of \( y \) into our differentiated equation: \[ \cos x=\frac{dy}{dx}\sin(x+a)+\frac{\sin x}{\sin(x+a)}\cos(x+a). \] Multiply through by \(\sin(x+a)\) to eliminate the denominator: \[ \cos x \sin(x+a)=\frac{dy}{dx}\sin^2(x+a)+\sin x \cos(x+a). \] Rearranging the equation to solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx}\sin^2(x+a)=\cos x \sin(x+a)-\sin x \cos(x+a). \] Recognize that the right-hand side is the sine of the difference of angles: \[ \cos x \sin(x+a)-\sin x \cos(x+a)=\sin\bigl((x+a)-x\bigr)=\sin a. \] Therefore, we obtain: \[ \frac{dy}{dx}\sin^2(x+a)=\sin a. \] Finally, solving for \(\frac{dy}{dx}\) gives: \[ \frac{dy}{dx}=\frac{\sin a}{\sin^2(x+a)}. \]

Solution: Start with the given equation: \[ \sin y = x \sin (a+y). \] Differentiate both sides with respect to \( x \). For the left-hand side, applying the chain rule: \[ \frac{d}{dx}(\sin y)=\cos y \cdot \frac{dy}{dx}. \] For the right-hand side, using the product rule: \[ \frac{d}{dx}\bigl(x \sin (a+y)\bigr)=\sin (a+y)+x\cos (a+y)\cdot\frac{dy}{dx}. \] Thus, we have: \[ \cos y\frac{dy}{dx} = \sin (a+y)+x\cos (a+y)\frac{dy}{dx}. \] Bringing the terms involving \(\frac{dy}{dx}\) to one side: \[ \cos y\frac{dy}{dx} – x\cos (a+y)\frac{dy}{dx} = \sin (a+y). \] Factor out \(\frac{dy}{dx}\): \[ \frac{dy}{dx}\left(\cos y – x\cos (a+y)\right)=\sin (a+y). \] Therefore, \[ \frac{dy}{dx}=\frac{\sin (a+y)}{\cos y – x\cos (a+y)}. \] From the original equation, express \( x \) as: \[ x=\frac{\sin y}{\sin (a+y)}. \] Substitute this value into the denominator: \[ \cos y – \frac{\sin y}{\sin (a+y)}\cos (a+y)=\frac{\cos y\sin (a+y)-\sin y\cos (a+y)}{\sin (a+y)}. \] Recognize that the numerator is the sine of a difference: \[ \cos y\sin (a+y)-\sin y\cos (a+y)=\sin\bigl((a+y)-y\bigr)=\sin a. \] Thus, the denominator simplifies to: \[ \frac{\sin a}{\sin (a+y)}. \] Substituting back, we obtain: \[ \frac{dy}{dx}=\frac{\sin (a+y)}{\frac{\sin a}{\sin (a+y)}}=\frac{\sin^2 (a+y)}{\sin a}. \]

Solution: Notice that the expression for \( y \) is self-referential. We can write: \[ y=\sqrt{x+y}. \] Squaring both sides gives: \[ y^{2}=x+y. \] Differentiate both sides with respect to \( x \): \[ 2y\frac{dy}{dx}=1+\frac{dy}{dx}. \] Rearranging to isolate \(\frac{dy}{dx}\): \[ 2y\frac{dy}{dx}-\frac{dy}{dx}=1, \] \[ (2y-1)\frac{dy}{dx}=1. \]

Solution: Notice that the infinite nested radical is self-referential, so we can write: \[ y=\sqrt{\sin x+y}. \] Squaring both sides gives: \[ y^2=\sin x+y. \] Differentiating both sides with respect to \( x \) yields: \[ 2y\frac{dy}{dx}=\cos x+\frac{dy}{dx}. \] Rearranging the terms to collect \(\frac{dy}{dx}\): \[ 2y\frac{dy}{dx}-\frac{dy}{dx}=\cos x, \] \[ (2y-1)\frac{dy}{dx}=\cos x. \]

Solution: Since the expression is infinitely nested and self-referential, we can set: \[ y=\sqrt{\tan x+y}. \] Squaring both sides: \[ y^2 = \tan x + y. \] Differentiate both sides with respect to \( x \): \[ 2y\frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}. \] Bringing like terms together: \[ 2y\frac{dy}{dx} – \frac{dy}{dx} = \sec^2 x, \] \[ (2y – 1)\frac{dy}{dx} = \sec^2 x. \]