Ex 5.4 class 12 maths ncert solutions | class 12 maths exercise 5.4 | class 12 maths ncert solutions chapter 5 exercise 5.4 | exercise 5.4 class 12 maths ncert solutions | continuity and differentiability class 12 ncert solutions
Looking for Ex 5.4 Class 12 Maths NCERT Solutions? You’re in the right place! This section provides step-by-step solutions for all problems from Exercise 5.4 Class 12 Maths, a part of Chapter 5 – Continuity and Differentiability. These NCERT solutions are crafted according to the latest CBSE syllabus to help students master concepts like the chain rule, derivatives of inverse trigonometric functions, and more. Whether you’re revising before exams or looking for clarity on complex topics, the Class 12 Maths Exercise 5.4 NCERT Solutions offer clear explanations and practice support. Strengthen your understanding of continuity and differentiability and boost your performance in board exams today!
exercise 5.4 class 12 maths ncert solutions || class 12 maths exercise 5.4 || ex 5.4 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 5 exercise 5.4 || continuity and differentiability class 12 ncert solutions
Exercise 5.4
1 . Differentiate the following w.r.t. \( x: \frac{e^{x}}{\sin x} \)
Answer
Let \( y=\frac{e^{x}}{\sin x} \)
By using the quotient rule, we get
\(\frac{d y}{d x}=\frac{\sin x \frac{d}{d x}\left(e^{x}\right)-e^{x} \frac{d}{d x}(\sin x)}{\sin ^{2} x}\)
\(=\frac{\sin x \cdot e^{x}-e^{x} \cdot(\cos x)}{\sin ^{2} x}\)
\(=\frac{e^{x}(\sin x-\cos x)}{\sin ^{2} x}\)
2 . Differentiate the following w.r.t. \( x \) :
\(e^{\sin ^{-1}} x\)
\(e^{\sin ^{-1}} x\)
Answer
Let \( y=e^{\sin ^{-1}} x \)
Now, by using the chain rule, we get,
\(\frac{d y}{d x}=\frac{d}{d x}\left(e^{\sin ^{-1}} x\right)\)
\(=\frac{d y}{d x}=e^{\sin ^{-1}} x \cdot \frac{d}{d x}\left(\sin ^{-1} x\right)\)
\(=e^{\sin ^{-1}} x \cdot \frac{21}{\sqrt{1-x^{2}}}\)
\(=\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^{2}}}\)
Thus, \( \frac{d y}{d x}=\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^{2}}} \)
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3 . Differentiate the following w.r.t. x:
\(e^{x^{3}}\)
\(e^{x^{3}}\)
Answer
Let \( y=e^{x^{3}} \)
So, by using the chain rule, we get,
\(\frac{d y}{d x}=\frac{d}{d x}\left(e^{x^{3}}\right)\)
\(=e^{x^{3}} \cdot \frac{d}{d x}\left(x^{3}\right)\)
\(=e^{x^{3}} \cdot 3 x^{2}\)
\(=3 x^{2} e^{x^{3}}\)
4 . Differentiate the following w.r.t. \( x \) :
\(\sin \left(\tan ^{-1} e^{-x}\right)\)
\(\sin \left(\tan ^{-1} e^{-x}\right)\)
Answer
Let \( y=\sin \left(\tan ^{-1} \mathrm{e}^{x}\right) \)
So, by using chain rule, we get
\(\frac{d y}{d x}=\frac{d}{d x}\left[\sin \left(\tan ^{-1} \mathrm{e}^{-x}\right)\right]\)
\(=\cos \left(\tan ^{-1} \mathrm{e}^{-x}\right) \cdot \frac{d}{d x}\left(\tan ^{-1} \mathrm{e}^{-x}\right)\)
\(=\cos \left(\tan ^{-1} \mathrm{e}^{-x}\right) \cdot \frac{1}{1+\left(e^{-x}\right)^{2}} \cdot \frac{d}{d x}\left(\mathrm{e}^{-x}\right)\)
\(=\frac{\cos \left(\tan ^{-1} e^{-x}\right)}{1+e^{-2 x}} \cdot \mathrm{e}-x \cdot \frac{d}{d x}(-x)\)
\(=\frac{e^{-x} \cos \left(\tan ^{-1} e^{-x}\right)}{1+e^{-2 x}} \cdot(-1)\)
\(=\frac{-e^{-x} \cos \left(\tan ^{-1} e^{-x}\right)}{1+e^{-2 x}}\)
5 . Differentiate the following w.r.t. \(x\):
\( \log \left(\cos \mathrm{e}^{x}\right) \)
\( \log \left(\cos \mathrm{e}^{x}\right) \)
Answer
Let \( y=\log \left(\cos \mathrm{e}^{x}\right) \)
So, by using the chain rule, we get,
\(\frac{d y}{d x}=\frac{d}{d x}\left(\log \left(\cos \mathrm{e}^{x}\right)\right)\)
\(=\frac{1}{\cos e^{x}} \cdot \frac{d}{d x}\left(\cos e^{x}\right)\)
\(=\frac{1}{\cos e^{x}} \cdot\left(-\sin e^{x}\right) \cdot \frac{d}{d x}\left(e^{x}\right)\)
\(=\frac{-\sin e^{x}}{\cos e^{x}} \cdot e^{x}\)
\(=-e^{x} \tan e^{x}\)
6 . Differentiate the following w.r.t. \( x \) :
\(e^{x}+e^{x^{2}}+\cdots+e^{x^{5}}\)
\(e^{x}+e^{x^{2}}+\cdots+e^{x^{5}}\)
Answer
\(\text { Let } y=e^{x}+e^{x^{2}}+\cdots+e^{x^{5}}\)
\(=\frac{d}{d x}\left(e^{x}+e^{x^{2}}+\cdots+e^{x^{5}}\right)\)
\(=\frac{d}{d x}\left(e^{x}\right)+\frac{d}{d x}\left(e^{x^{2}}\right)+\frac{d}{d x}\left(e^{x^{3}}\right)+\frac{d}{d x}\left(e^{x^{4}}\right)+\frac{d}{d x}\left(e^{x^{5}}\right)\)
\(=e^{x}+e^{x^{2}} \cdot 2 x+e^{x^{3}} \cdot 3 x^{2}+e^{x^{4}} \cdot 4 x^{3}+e^{x^{5}} \cdot 5 x^{4}\)
\(=e^{x}+2 x e^{x^{2}}+3 x^{2} e^{x^{3}}+4 x^{3} e^{x^{4}}+5 x^{4} e^{x^{5}}\)
7 . Differentiate the following w.r.t. \(x\):
\(\sqrt{e^{\sqrt{x}}}, x > 0\)
\(\sqrt{e^{\sqrt{x}}}, x > 0\)
Answer
Let \( y=\sqrt{e^{\sqrt{x}}} \)
Then, \( y^{2}=e^{\sqrt{x}} \)
Now, differentiating both sides we get,
\(2 y \frac{d y}{d x}=e^{\sqrt{x}} \frac{d}{d x}(\sqrt{x})\)
\(=e^{\sqrt{x}} \frac{1}{2} \cdot \frac{1}{\sqrt{x}}\)
\(=\frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 y \sqrt{x}}\)
\(=\frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{e^{\sqrt{x}}} \sqrt{x}}\)
\(=\frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{x e^{\sqrt{x}}}}\)
exercise 5.4 class 12 maths ncert solutions || class 12 maths exercise 5.4 || ex 5.4 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 5 exercise 5.4 || continuity and differentiability class 12 ncert solutions
8 . Differentiate the following w.r.t. \( x \) :
\(\log^{(\log x)}, x > 1\)
\(\log^{(\log x)}, x > 1\)
Answer
\(\text { let } y=\log (\log x)\)
So, by using chain rule, we get,
\(\frac{d y}{d x}=\frac{d}{d x}(\log (\log x))\)
\(=\frac{1}{\log x} \cdot \frac{d}{d x}(\log x)\)
\(=\frac{1}{\log x} \cdot \frac{1}{x}\)
\(=\frac{1}{x \log x}\)
9 . Differentiate the following w.r.t. \(x\):
\(\frac{\cos x}{\log x}, x > 0\)
\(\frac{\cos x}{\log x}, x > 0\)
Answer
Let \( y=\frac{\cos x}{\log x}, x > 0 \)
So, by using the quotient rule, we get,
\(\frac{d y}{d x}=\frac{\frac{d}{d x}(\cos x) \times \log x-\cos x \times \frac{d}{d x}(\log x)}{(\log x)^{2}}\)
\(=\frac{-\sin x \log x-\cos x \times \frac{1}{x}}{(\log x)^{2}}\)
\(=\frac{-[x \log x \cdot \sin x+\cos x]}{x(\log x)^{2}}\)
10. Differentiate the following w.r.t. \(x \):
\(\cos \left(\log x+\mathrm{e}^{x}\right), x > 0\)
\(\cos \left(\log x+\mathrm{e}^{x}\right), x > 0\)
Answer
Let \( y=\cos \left(\log x+\mathrm{e}^{x}\right) \)
So, by using chain rule, we get,
\(\frac{d y}{d x}=-\sin \left(\log x+e^{x}\right) \cdot \frac{d}{d x}\left(\log x+e^{x}\right)\)
\(=-\sin \left(\log x+e^{x} \cdot\left[\frac{d}{d x}(\log x)+\frac{d}{d x}\left(e^{x}\right)\right]\right.\)
\(=-\sin \left(\log x+e^{x}\right) \cdot\left(\frac{1}{x}+e^{x}\right)\)
\(=-\left(\frac{1}{x}+e^{x}\right) \sin \left(\log x+e^{x}\right)\)
