Ex 5.6 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution:

We have \( f(x)=\sqrt{4-x^2}=(4-x^2)^{\frac{1}{2}} \).

Differentiating using the chain rule: \[ f'(x)=\frac{1}{2}(4-x^2)^{-\frac{1}{2}}(-2x)=-\frac{x}{\sqrt{4-x^2}}. \]

Solution:

We have \( f(x)=\sqrt{1-x^2}=(1-x^2)^{\frac{1}{2}} \).

Differentiating using the chain rule: \[ f'(x)=\frac{1}{2}(1-x^2)^{-\frac{1}{2}}(-2x)=-\frac{x}{\sqrt{1-x^2}}. \]

Solution:

Write the function as \( \sqrt{2x-1}=(2x-1)^{\frac{1}{2}} \).

Using the chain rule, differentiate to get: \[ \frac{d}{dx}\left((2x-1)^{\frac{1}{2}}\right)=\frac{1}{2}(2x-1)^{-\frac{1}{2}} \cdot 2 = \frac{1}{\sqrt{2x-1}}. \]

Solution:

Let \( f(x)=\left(3x^{2}-9x+5\right)^{9} \).

Applying the chain rule, we differentiate the outer function and then multiply by the derivative of the inner function.

The derivative is: \[ f'(x)=9\left(3x^{2}-9x+5\right)^{8}\cdot\frac{d}{dx}(3x^{2}-9x+5). \]

Now, differentiating the inner function: \[ \frac{d}{dx}(3x^{2}-9x+5)=6x-9. \]

Thus, the derivative becomes: \[ f'(x)=9\left(3x^{2}-9x+5\right)^{8}(6x-9). \]

Solution:

Write the first term as \( (3x+2)^{\frac{1}{2}} \). Differentiating, we get: \[ \frac{d}{dx}(3x+2)^{\frac{1}{2}}=\frac{1}{2}(3x+2)^{-\frac{1}{2}} \cdot 3=\frac{3}{2\sqrt{3x+2}}. \]

Write the second term as \( (2x^2+4)^{-\frac{1}{2}} \). Differentiating, we have: \[ \frac{d}{dx}(2x^2+4)^{-\frac{1}{2}}=-\frac{1}{2}(2x^2+4)^{-\frac{3}{2}} \cdot (4x)=-\frac{2x}{(2x^2+4)^{\frac{3}{2}}}. \]

Therefore, the derivative of the given function is: \[ \frac{d}{dx}\left(\sqrt{3x+2}+\frac{1}{\sqrt{2x^2+4}}\right)=\frac{3}{2\sqrt{3x+2}}-\frac{2x}{(2x^2+4)^{\frac{3}{2}}}. \]

Solution:

Express the terms as powers of trigonometric functions:
\[ \sin^3 x = (\sin x)^3 \quad \text{and} \quad \cos^6 x = (\cos x)^6. \]

Differentiating the first term using the chain rule: \[ \frac{d}{dx}(\sin x)^3 = 3(\sin x)^2 \cdot \cos x. \]

Differentiating the second term: \[ \frac{d}{dx}(\cos x)^6 = 6(\cos x)^5 \cdot (-\sin x) = -6(\cos x)^5 \sin x. \]

Combining the derivatives, we obtain: \[ \frac{d}{dx}\left(\sin^3 x + \cos^6 x\right) = 3\sin^2 x \cos x – 6\cos^5 x \sin x. \]

Solution:

Let \( f(x) = \sin(x^2) \).

Using the chain rule, we differentiate the outer function \( \sin(u) \) where \( u = x^2 \): \[ \frac{d}{dx}(\sin(x^2)) = \cos(x^2) \cdot \frac{d}{dx}(x^2). \]

Now, \[ \frac{d}{dx}(x^2) = 2x. \]

So the final derivative becomes: \[ \frac{d}{dx}(\sin(x^2)) = 2x \cos(x^2). \]

Solution:

Let \( f(x) = \cos(\sqrt{x}) \).

Using the chain rule, we differentiate the outer function \( \cos(u) \) where \( u = \sqrt{x} \): \[ \frac{d}{dx}(\cos(\sqrt{x})) = -\sin(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}). \]

Now, \[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}. \]

So the derivative becomes: \[ \frac{d}{dx}(\cos(\sqrt{x})) = -\frac{\sin(\sqrt{x})}{2\sqrt{x}}. \]

Solution:

Let \( f(x) = \sin(ax + b) \).

Using the chain rule, the derivative of \( \sin(u) \) where \( u = ax + b \) is: \[ \frac{d}{dx}(\sin(ax + b)) = \cos(ax + b) \cdot \frac{d}{dx}(ax + b). \]

Since, \[ \frac{d}{dx}(ax + b) = a, \] the final derivative becomes: \[ \frac{d}{dx}(\sin(ax + b)) = a \cos(ax + b). \]

Solution:

Let \( f(x) = \tan(2x + 3) \).

Using the chain rule, we differentiate the outer function \( \tan(u) \) where \( u = 2x + 3 \): \[ \frac{d}{dx}(\tan(2x + 3)) = \sec^2(2x + 3) \cdot \frac{d}{dx}(2x + 3). \]

Now, \[ \frac{d}{dx}(2x + 3) = 2. \]

So the derivative becomes: \[ \frac{d}{dx}(\tan(2x + 3)) = 2 \sec^2(2x + 3). \]

Solution:

Let \( f(x) = \cos(\sin x) \).

Using the chain rule, the derivative of \( \cos(u) \) where \( u = \sin x \) is: \[ \frac{d}{dx}(\cos(\sin x)) = -\sin(\sin x) \cdot \frac{d}{dx}(\sin x). \]

Now, \[ \frac{d}{dx}(\sin x) = \cos x. \]

So the derivative becomes: \[ \frac{d}{dx}(\cos(\sin x)) = -\sin(\sin x) \cdot \cos x. \]

Solution:

Let \( f(x) = \sin(\cos(x^2)) \).

Using the chain rule, we differentiate the outer function \( \sin(u) \) where \( u = \cos(x^2) \): \[ \frac{d}{dx}(\sin(\cos(x^2))) = \cos(\cos(x^2)) \cdot \frac{d}{dx}(\cos(x^2)). \]

Now, \[ \frac{d}{dx}(\cos(x^2)) = -\sin(x^2) \cdot \frac{d}{dx}(x^2) = -\sin(x^2) \cdot 2x. \]

So the derivative becomes: \[ \frac{d}{dx}(\sin(\cos(x^2))) = \cos(\cos(x^2)) \cdot (-\sin(x^2) \cdot 2x). \]

Simplifying: \[ \frac{d}{dx}(\sin(\cos(x^2))) = -2x \sin(x^2) \cos(\cos(x^2)). \]

Solution:

Let \( f(x) = \cos\left(\tan\sqrt{x + 1}\right) \).

Using the chain rule: \[ \frac{d}{dx}\left[\cos\left(\tan\sqrt{x + 1}\right)\right] = -\sin\left(\tan\sqrt{x + 1}\right) \cdot \frac{d}{dx}\left(\tan\sqrt{x + 1}\right). \]

Now differentiate \( \tan\sqrt{x + 1} \): \[ \frac{d}{dx}(\tan\sqrt{x + 1}) = \sec^2(\sqrt{x + 1}) \cdot \frac{d}{dx}(\sqrt{x + 1}). \]
\[ \frac{d}{dx}(\sqrt{x + 1}) = \frac{1}{2\sqrt{x + 1}}. \]

Putting it all together: \[ \frac{d}{dx}\left[\cos\left(\tan\sqrt{x + 1}\right)\right] = -\sin\left(\tan\sqrt{x + 1}\right) \cdot \sec^2(\sqrt{x + 1}) \cdot \frac{1}{2\sqrt{x + 1}}. \]

Solution:

Let \( f(x) = \sqrt{\tan \sqrt{x}} = (\tan \sqrt{x})^{1/2} \).

Using the chain rule: \[ \frac{d}{dx} \left( (\tan \sqrt{x})^{1/2} \right) = \frac{1}{2} (\tan \sqrt{x})^{-1/2} \cdot \frac{d}{dx} (\tan \sqrt{x}). \]
Now, \[ \frac{d}{dx}(\tan \sqrt{x}) = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}. \]
So, \[ \frac{d}{dx} \left( \sqrt{\tan \sqrt{x}} \right) = \frac{1}{2\sqrt{\tan \sqrt{x}}} \cdot \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}. \]
Simplifying: \[ \frac{d}{dx} \left( \sqrt{\tan \sqrt{x}} \right) = \frac{\sec^2(\sqrt{x})}{4\sqrt{x} \cdot \sqrt{\tan \sqrt{x}}}. \]

Solution:

The function is: \[ f(x) = \sum_{k=1}^{100} \frac{x^k}{k} + 1 \]
Differentiating term-by-term: \[ f'(x) = \sum_{k=1}^{100} \frac{d}{dx} \left( \frac{x^k}{k} \right) = \sum_{k=1}^{100} \frac{kx^{k-1}}{k} = \sum_{k=1}^{100} x^{k-1} \]
So, \[ f'(x) = x^0 + x^1 + x^2 + \ldots + x^{99} \]
Now evaluate \( f'(1) \) and \( f'(0) \):

\[ f'(1) = 1 + 1 + 1 + \ldots + 1 \quad (\text{100 terms}) = 100 \]
\[ f'(0) = 0^0 + 0^1 + 0^2 + \ldots + 0^{99} = 1 + 0 + 0 + \ldots + 0 = 1 \]
Hence, \[ f'(1) = 100 = 100 \cdot 1 = 100f'(0) \]

Solution:

Let \( f(x) = |2x^2 – 3| \). To differentiate the absolute value function, we consider the definition: \[ \frac{d}{dx} |u(x)| = \frac{u(x)}{|u(x)|} \cdot u'(x), \quad \text{for } u(x) \ne 0 \]
Here, \( u(x) = 2x^2 – 3 \) and \( u'(x) = 4x \). So, \[ \frac{d}{dx} \left|2x^2 – 3\right| = \frac{2x^2 – 3}{|2x^2 – 3|} \cdot 4x, \quad \text{for } x \ne \pm \sqrt{\frac{3}{2}} \]

Solution:

Let \( f(x) = |\cos x| \). To differentiate, use the identity: \[ \frac{d}{dx}|\cos x| = \frac{\cos x}{|\cos x|} \cdot (-\sin x), \quad \text{for } \cos x \ne 0 \]
Now evaluate at \( x = \frac{3\pi}{4} \):
\[ \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}, \quad \sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
So, \[ f’\left(\frac{3\pi}{4}\right) = \frac{-\frac{1}{\sqrt{2}}}{\left| -\frac{1}{\sqrt{2}} \right|} \cdot (-\frac{1}{\sqrt{2}}) = (-1) \cdot \left(-\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} \]

Solution:

Let \[ f(x) = \sqrt{\frac{a^2 – x^2}{a^2 + x^2}} = \left( \frac{a^2 – x^2}{a^2 + x^2} \right)^{1/2} \] Using the chain rule: \[ \frac{d}{dx} f(x) = \frac{1}{2} \left( \frac{a^2 – x^2}{a^2 + x^2} \right)^{-1/2} \cdot \frac{d}{dx} \left( \frac{a^2 – x^2}{a^2 + x^2} \right) \] Let us differentiate the inner function using the quotient rule: \[ \frac{d}{dx} \left( \frac{a^2 – x^2}{a^2 + x^2} \right) = \frac{(a^2 + x^2)(-2x) – (a^2 – x^2)(2x)}{(a^2 + x^2)^2} \] Simplifying the numerator: \[ = -2x(a^2 + x^2) – 2x(a^2 – x^2) = -2x[(a^2 + x^2) + (a^2 – x^2)] = -2x(2a^2) = -4a^2x \] So the derivative becomes: \[ f'(x) = \frac{1}{2} \left( \frac{a^2 – x^2}{a^2 + x^2} \right)^{-1/2} \cdot \frac{-4a^2x}{(a^2 + x^2)^2} \]

Solution:

Let \[ f(x) = \sin \sqrt{x} + \cos^2 \sqrt{x} \] Differentiate term by term: 1. For \( \sin \sqrt{x} \), Let \( u = \sqrt{x} \), so \( \frac{d}{dx} \sin u = \cos u \cdot \frac{du}{dx} = \cos \sqrt{x} \cdot \frac{1}{2\sqrt{x}} \) 2. For \( \cos^2 \sqrt{x} \), Use the chain rule: \[ \frac{d}{dx} (\cos \sqrt{x})^2 = 2\cos \sqrt{x} \cdot \frac{d}{dx} (\cos \sqrt{x}) = 2\cos \sqrt{x} \cdot (-\sin \sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \] Combine both derivatives: \[ f'(x) = \frac{\cos \sqrt{x}}{2\sqrt{x}} + 2\cos \sqrt{x} \cdot (-\sin \sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \] \[ = \frac{\cos \sqrt{x}}{2\sqrt{x}} – \frac{\sin \sqrt{x} \cdot \cos \sqrt{x}}{\sqrt{x}} \]

Solution:

Let \[ f(x) = \frac{\sin(ax + b)}{\cos(cx + d)} \] Use the quotient rule: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} – u \cdot \frac{dv}{dx}}{v^2} \] Here, \( u = \sin(ax + b) \Rightarrow \frac{du}{dx} = a \cos(ax + b) \) \( v = \cos(cx + d) \Rightarrow \frac{dv}{dx} = -c \sin(cx + d) \) Substitute into the quotient rule: \[ f'(x) = \frac{\cos(cx + d) \cdot a \cos(ax + b) – \sin(ax + b) \cdot (-c \sin(cx + d))}{\cos^2(cx + d)} \] \[ = \frac{a \cos(ax + b) \cos(cx + d) + c \sin(ax + b) \sin(cx + d)}{\cos^2(cx + d)} \]

Solution:

Let \[ f(x) = \frac{\sin x + x^2}{\cot 2x} \] Use the quotient rule: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} – u \cdot \frac{dv}{dx}}{v^2} \] Here, \( u = \sin x + x^2 \Rightarrow \frac{du}{dx} = \cos x + 2x \) \( v = \cot 2x \Rightarrow \frac{dv}{dx} = -2 \csc^2 2x \) Substitute into the formula: \[ f'(x) = \frac{\cot 2x \cdot (\cos x + 2x) – (\sin x + x^2) \cdot (-2 \csc^2 2x)}{(\cot 2x)^2} \] \[ = \frac{\cot 2x (\cos x + 2x) + 2(\sin x + x^2)\csc^2 2x}{\cot^2 2x} \]

Solution:

Let \[ f(x) = \sin^m x \cdot \cos^n x \] Apply the product rule: \[ \frac{d}{dx}(u \cdot v) = u’v + uv’ \] Here, \( u = \sin^m x \Rightarrow \frac{du}{dx} = m \sin^{m-1} x \cdot \cos x \) \( v = \cos^n x \Rightarrow \frac{dv}{dx} = n \cos^{n-1} x \cdot (-\sin x) \) Now, \[ f'(x) = m \sin^{m-1} x \cos x \cdot \cos^n x + \sin^m x \cdot n \cos^{n-1} x \cdot (-\sin x) \] \[ = m \sin^{m-1} x \cos^{n+1} x – n \sin^{m+1} x \cos^{n-1} x \]

Solution:

Let \[ f(x) = \sin^n(ax^2 + bx + c) \] This is a composite function, so apply the chain rule: \[ \frac{d}{dx}[\sin^n(u)] = n \sin^{n-1}(u) \cdot \cos(u) \cdot \frac{du}{dx} \] Here, \( u = ax^2 + bx + c \Rightarrow \frac{du}{dx} = 2ax + b \) Therefore, \[ f'(x) = n \sin^{n-1}(ax^2 + bx + c) \cdot \cos(ax^2 + bx + c) \cdot (2ax + b) \]

Solution:

Let \[ f(x) = \frac{\sqrt{a+x} – \sqrt{a-x}}{\sqrt{a+x} + \sqrt{a-x}} \] Apply the quotient rule: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u’ – u \cdot v’}{v^2} \] Let \( u = \sqrt{a+x} – \sqrt{a-x} \Rightarrow u’ = \frac{1}{2\sqrt{a+x}} + \frac{1}{2\sqrt{a-x}} \) \( v = \sqrt{a+x} + \sqrt{a-x} \Rightarrow v’ = \frac{1}{2\sqrt{a+x}} – \frac{1}{2\sqrt{a-x}} \) Now, \[ f'(x) = \frac{(\sqrt{a+x} + \sqrt{a-x}) \left( \frac{1}{2\sqrt{a+x}} + \frac{1}{2\sqrt{a-x}} \right) – (\sqrt{a+x} – \sqrt{a-x}) \left( \frac{1}{2\sqrt{a+x}} – \frac{1}{2\sqrt{a-x}} \right)}{(\sqrt{a+x} + \sqrt{a-x})^2} \] Simplify the numerator: \[ = \frac{1 + 1 + 1 + 1}{(\sqrt{a+x} + \sqrt{a-x})^2} = \frac{2}{(\sqrt{a+x} + \sqrt{a-x})^2} \]

Solution:

Let \[ f(x) = \frac{\sqrt{x^2 + 1} + \sqrt{x^2 – 1}}{\sqrt{x^2 + 1} – \sqrt{x^2 – 1}} \] Use the quotient rule: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u’ – u \cdot v’}{v^2} \] Let \( u = \sqrt{x^2 + 1} + \sqrt{x^2 – 1} \Rightarrow u’ = \frac{x}{\sqrt{x^2 + 1}} + \frac{x}{\sqrt{x^2 – 1}} \) \( v = \sqrt{x^2 + 1} – \sqrt{x^2 – 1} \Rightarrow v’ = \frac{x}{\sqrt{x^2 + 1}} – \frac{x}{\sqrt{x^2 – 1}} \) Now apply the quotient rule: \[ f'(x) = \frac{(\sqrt{x^2 + 1} – \sqrt{x^2 – 1}) \left( \frac{x}{\sqrt{x^2 + 1}} + \frac{x}{\sqrt{x^2 – 1}} \right) – (\sqrt{x^2 + 1} + \sqrt{x^2 – 1}) \left( \frac{x}{\sqrt{x^2 + 1}} – \frac{x}{\sqrt{x^2 – 1}} \right)}{\left( \sqrt{x^2 + 1} – \sqrt{x^2 – 1} \right)^2} \] While the simplification is a bit long, this is the correct differentiated expression.

Solution:

Let \[ f(x) = \sqrt{ \frac{ \sec x – 1 }{ \sec x + 1 } } = \left( \frac{ \sec x – 1 }{ \sec x + 1 } \right)^{1/2} \] Using the chain rule: \[ f'(x) = \frac{1}{2} \left( \frac{ \sec x – 1 }{ \sec x + 1 } \right)^{-1/2} \cdot \frac{d}{dx} \left( \frac{ \sec x – 1 }{ \sec x + 1 } \right) \] Let \( u = \sec x – 1 \), \( v = \sec x + 1 \) Then, \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u’ – u \cdot v’}{v^2} \] Now, \( u’ = \frac{d}{dx}(\sec x – 1) = \sec x \tan x \) \( v’ = \frac{d}{dx}(\sec x + 1) = \sec x \tan x \) So, \[ \frac{d}{dx} \left( \frac{ \sec x – 1 }{ \sec x + 1 } \right) = \frac{ (\sec x + 1)(\sec x \tan x) – (\sec x – 1)(\sec x \tan x) }{ (\sec x + 1)^2 } \] Simplifying numerator: \[ \sec x \tan x [ (\sec x + 1) – (\sec x – 1) ] = \sec x \tan x \cdot 2 \] So, \[ \frac{d}{dx} \left( \frac{ \sec x – 1 }{ \sec x + 1 } \right) = \frac{ 2 \sec x \tan x }{ (\sec x + 1)^2 } \] Therefore, \[ f'(x) = \frac{1}{2} \left( \frac{ \sec x – 1 }{ \sec x + 1 } \right)^{-1/2} \cdot \frac{ 2 \sec x \tan x }{ (\sec x + 1)^2 } \] Cancelling 2: \[ f'(x) = \frac{ \sec x \tan x }{ (\sec x + 1)^2 \cdot \sqrt{ \frac{ \sec x – 1 }{ \sec x + 1 } } } \] Now compute \( f’\left( \frac{\pi}{3} \right) \): \[ \sec \left( \frac{\pi}{3} \right) = 2, \quad \tan \left( \frac{\pi}{3} \right) = \sqrt{3} \] Substitute into derivative: \[ f’\left( \frac{\pi}{3} \right) = \frac{ 2 \cdot \sqrt{3} }{ (2 + 1)^2 \cdot \sqrt{ \frac{2 – 1}{2 + 1} } } = \frac{ 2 \sqrt{3} }{ 9 \cdot \sqrt{ \frac{1}{3} } } = \frac{ 2 \sqrt{3} }{ 9 \cdot \frac{1}{\sqrt{3}} } = \frac{ 2 \sqrt{3} \cdot \sqrt{3} }{ 9 } = \frac{ 6 }{ 9 } = \frac{2}{3} \]

Solution:

Given: \[ y = x \sin^{-1} x + \sqrt{1 – x^2} \] Differentiate both terms using the product rule and chain rule: First term: \[ \frac{d}{dx}(x \sin^{-1} x) = \frac{d}{dx}(x) \cdot \sin^{-1} x + x \cdot \frac{d}{dx}(\sin^{-1} x) \] \[ = 1 \cdot \sin^{-1} x + x \cdot \frac{1}{\sqrt{1 – x^2}} = \sin^{-1} x + \frac{x}{\sqrt{1 – x^2}} \] Second term: \[ \frac{d}{dx}(\sqrt{1 – x^2}) = \frac{1}{2 \sqrt{1 – x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1 – x^2}} \] Combine both derivatives: \[ \frac{dy}{dx} = \left( \sin^{-1} x + \frac{x}{\sqrt{1 – x^2}} \right) + \left( \frac{-x}{\sqrt{1 – x^2}} \right) \] Simplify: \[ \frac{dy}{dx} = \sin^{-1} x \]

Solution:

Given: \[ y = \sqrt{x} + \frac{1}{\sqrt{x}} = x^{1/2} + x^{-1/2} \] Differentiate with respect to \( x \): \[ \frac{dy}{dx} = \frac{1}{2}x^{-1/2} – \frac{1}{2}x^{-3/2} \] Multiply both sides by \( 2x \): \[ 2x \cdot \frac{dy}{dx} = 2x \left( \frac{1}{2}x^{-1/2} – \frac{1}{2}x^{-3/2} \right) \] Simplify: \[ 2x \cdot \frac{dy}{dx} = x^{1/2} – x^{-1/2} \] Now add \( y \) to both sides: \[ 2x \frac{dy}{dx} + y = \left( x^{1/2} – x^{-1/2} \right) + \left( x^{1/2} + x^{-1/2} \right) \] Combine like terms: \[ 2x \frac{dy}{dx} + y = 2x^{1/2} \] That is: \[ 2x \frac{dy}{dx} + y = 2\sqrt{x} \]

Solution:

Given: \[ y = \sqrt{\frac{1 – \sin 2x}{1 + \sin 2x}} \] This can be rewritten using the identity: \[ \frac{1 – \sin 2x}{1 + \sin 2x} = \tan^2\left( \frac{\pi}{4} – x \right) \] So, \[ y = \sqrt{\tan^2\left( \frac{\pi}{4} – x \right)} = \left| \tan\left( \frac{\pi}{4} – x \right) \right| \] Since \( \frac{\pi}{4} – x \in \left(0, \frac{\pi}{2}\right) \) for small \( x \), \( \tan\left( \frac{\pi}{4} – x \right) > 0 \), so: \[ y = \tan\left( \frac{\pi}{4} – x \right) \] Differentiate: \[ \frac{dy}{dx} = \frac{d}{dx} \left( \tan\left( \frac{\pi}{4} – x \right) \right) \] \[ = \sec^2\left( \frac{\pi}{4} – x \right) \cdot (-1) \] \[ = -\sec^2\left( \frac{\pi}{4} – x \right) \] Add \( \sec^2\left( \frac{\pi}{4} – x \right) \) to both sides: \[ \frac{dy}{dx} + \sec^2\left( \frac{\pi}{4} – x \right) = 0 \]

Solution:

First, consider the function \( f(x) = |\cos x| \). The absolute value function is defined as: \[ |\cos x| = \begin{cases} \cos x, & \text{if } \cos x \geq 0 \\ -\cos x, & \text{if } \cos x < 0 \end{cases} \] Differentiating case-wise: - When \( \cos x > 0 \), \[ \frac{d}{dx}|\cos x| = \frac{d}{dx}(\cos x) = -\sin x \] – When \( \cos x < 0 \), \[ \frac{d}{dx}|\cos x| = \frac{d}{dx}(-\cos x) = \sin x \] - When \( \cos x = 0 \), i.e., at points \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots \), the left-hand derivative and right-hand derivative are not equal, so the function is **not differentiable** at those points. Now consider \( g(x) = \cos |x| \). Since \( |x| \) is differentiable for all \( x \neq 0 \), and \( \cos x \) is differentiable everywhere, we have: \[ \cos |x| = \cos x \quad \text{for all real } x, \] and \( \cos x \) is differentiable for all \( x \). Therefore, \( \cos |x| \) is differentiable for all \( x \).