NCERT Solutions Class-12-Chapter-5-Differentiability and Continuity
Excercise-5.4
Note:
Exponential Derivative: \( \frac{d}{dx}(e^x) = e^x \). Using Chain Rule: \( \frac{d}{dx}(e^{f(x)}) = e^{f(x)} \cdot f'(x) \).
Logarithmic Derivative: \( \frac{d}{dx}(\log x) = \frac{1}{x} \). Using Chain Rule: \( \frac{d}{dx}(\log f(x)) = \frac{1}{f(x)} \cdot f'(x) \).
Q1
Differentiate the following w.r.t. \( x \): \( \frac{e^x}{\sin x} \)
Solution:
Using the Quotient Rule: \( \left( \frac{u}{v} \right)' = \frac{v u' - u v'}{v^2} \).
Here \( u = e^x \) and \( v = \sin x \).
\[ \begin{aligned} \frac{dy}{dx} &= \frac{\sin x \frac{d}{dx}(e^x) - e^x \frac{d}{dx}(\sin x)}{(\sin x)^2} \\ &= \frac{\sin x \cdot e^x - e^x \cdot \cos x}{\sin^2 x} \\ &= \frac{e^x(\sin x - \cos x)}{\sin^2 x} \end{aligned} \]
Q2
Differentiate the following w.r.t. \( x \): \( e^{\sin^{-1} x} \)
Solution:
Using the Chain Rule:
\[ \begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}(e^{\sin^{-1} x}) \\ &= e^{\sin^{-1} x} \cdot \frac{d}{dx}(\sin^{-1} x) \\ &= e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1 - x^2}} \\ &= \frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}} \end{aligned} \]
Q3
Differentiate the following w.r.t. \( x \): \( e^{x^3} \)
Solution:
Using the Chain Rule:
\[ \begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}(e^{x^3}) \\ &= e^{x^3} \cdot \frac{d}{dx}(x^3) \\ &= e^{x^3} \cdot 3x^2 \\ &= 3x^2 e^{x^3} \end{aligned} \]
Q4
Differentiate the following w.r.t. \( x \): \( \sin(\tan^{-1} e^{-x}) \)
Solution:
Apply Chain Rule sequentially:
Let \( y = \sin(\tan^{-1} e^{-x}) \).
\[ \begin{aligned} \frac{dy}{dx} &= \cos(\tan^{-1} e^{-x}) \cdot \frac{d}{dx}(\tan^{-1} e^{-x}) \\ &= \cos(\tan^{-1} e^{-x}) \cdot \frac{1}{1 + (e^{-x})^2} \cdot \frac{d}{dx}(e^{-x}) \\ &= \cos(\tan^{-1} e^{-x}) \cdot \frac{1}{1 + e^{-2x}} \cdot (e^{-x} \cdot -1) \\ &= -\frac{e^{-x} \cos(\tan^{-1} e^{-x})}{1 + e^{-2x}} \end{aligned} \]
Q5
Differentiate the following w.r.t. \( x \): \( \log(\cos e^x) \)
Solution:
Using Chain Rule:
\[ \begin{aligned} \frac{dy}{dx} &= \frac{1}{\cos e^x} \cdot \frac{d}{dx}(\cos e^x) \\ &= \frac{1}{\cos e^x} \cdot (-\sin e^x) \cdot \frac{d}{dx}(e^x) \\ &= \frac{-\sin e^x}{\cos e^x} \cdot e^x \\ &= -e^x \tan(e^x) \end{aligned} \]
Q6
Differentiate the following w.r.t. \( x \): \( e^x + e^{x^2} + \dots + e^{x^5} \)
Solution:
Differentiate each term individually:
\[ \begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}(e^x) + \frac{d}{dx}(e^{x^2}) + \frac{d}{dx}(e^{x^3}) + \frac{d}{dx}(e^{x^4}) + \frac{d}{dx}(e^{x^5}) \\ &= e^x + e^{x^2}(2x) + e^{x^3}(3x^2) + e^{x^4}(4x^3) + e^{x^5}(5x^4) \\ &= e^x + 2x e^{x^2} + 3x^2 e^{x^3} + 4x^3 e^{x^4} + 5x^4 e^{x^5} \end{aligned} \]
Q7
Differentiate the following w.r.t. \( x \): \( \sqrt{e^{\sqrt{x}}}, x > 0 \)
Solution:
Let \( y = \sqrt{e^{\sqrt{x}}} = (e^{\sqrt{x}})^{1/2} = e^{\frac{1}{2}\sqrt{x}} \).
Alternatively, using Chain Rule on the square root form:
\[ \begin{aligned} \frac{dy}{dx} &= \frac{1}{2\sqrt{e^{\sqrt{x}}}} \cdot \frac{d}{dx}(e^{\sqrt{x}}) \\ &= \frac{1}{2\sqrt{e^{\sqrt{x}}}} \cdot e^{\sqrt{x}} \cdot \frac{d}{dx}(\sqrt{x}) \\ &= \frac{e^{\sqrt{x}}}{2\sqrt{e^{\sqrt{x}}}} \cdot \frac{1}{2\sqrt{x}} \\ &= \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}} \end{aligned} \]
Q8
Differentiate the following w.r.t. \( x \): \( \log(\log x), x > 1 \)
Solution:
\[ \begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}(\log(\log x)) \\ &= \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) \\ &= \frac{1}{\log x} \cdot \frac{1}{x} \\ &= \frac{1}{x \log x} \end{aligned} \]
Q9
Differentiate the following w.r.t. \( x \): \( \frac{\cos x}{\log x}, x > 0 \)
Solution:
Using Quotient Rule:
\[ \begin{aligned} \frac{dy}{dx} &= \frac{\log x \frac{d}{dx}(\cos x) - \cos x \frac{d}{dx}(\log x)}{(\log x)^2} \\ &= \frac{\log x (-\sin x) - \cos x (\frac{1}{x})}{(\log x)^2} \\ &= \frac{-x \sin x \log x - \cos x}{x (\log x)^2} \\ &= -\frac{x \sin x \log x + \cos x}{x (\log x)^2} \end{aligned} \]
Q10
Differentiate the following w.r.t. \( x \): \( \cos(\log x + e^x), x > 0 \)
Solution:
Using Chain Rule:
\[ \begin{aligned} \frac{dy}{dx} &= -\sin(\log x + e^x) \cdot \frac{d}{dx}(\log x + e^x) \\ &= -\sin(\log x + e^x) \cdot \left( \frac{1}{x} + e^x \right) \\ &= -\left( \frac{1}{x} + e^x \right) \sin(\log x + e^x) \end{aligned} \]