Where Mindful Learning Begins
Where Mindful Learning Begins
This Class 12 Maths NCERT Solutions Chapter 5 Ex 5.7 page covers all 17 questions solved step-by-step — differentiating a function twice to find d^2y/dx^2, on polynomial, trigonometric, logarithmic and exponential functions, plus the proof-based questions that use implicit differentiation twice in a row.
This exercise splits cleanly into two halves. Questions 1–10 are direct second order derivative problems — you differentiate a given function once, then differentiate the result again, covering everything from simple polynomials like x^{20} to product-rule cases like x\cos x and x^3\log x, and mixed exponential-trig functions like e^x\sin 5x. Questions 11–17 flip the problem around: you're given a relationship between x and y and asked to prove a second-order differential equation holds — a pattern (differentiate, simplify, differentiate again) that shows up constantly in CBSE board papers and in Chapter 9 (Differential Equations) later in the year, which is exactly why getting comfortable with it here pays off well beyond this one exercise.
\dfrac{dy}{dx}=2x+3
\dfrac{d^2y}{dx^2}=2
\dfrac{dy}{dx}=20x^{19}
\dfrac{d^2y}{dx^2}=380x^{18}
Product rule: \dfrac{dy}{dx}=\cos x-x\sin x
Differentiate again (product rule on x\sin x): \dfrac{d^2y}{dx^2}=-\sin x-\big[\sin x+x\cos x\big]
\dfrac{dy}{dx}=\dfrac{1}{x}
\dfrac{d^2y}{dx^2}=-\dfrac{1}{x^2}
Product rule: \dfrac{dy}{dx}=3x^2\log x+x^2
Differentiate again: \dfrac{d^2y}{dx^2}=6x\log x+3x+2x
Product rule: \dfrac{dy}{dx}=e^x\sin 5x+5e^x\cos 5x=e^x(\sin 5x+5\cos 5x)
Differentiate again (product rule): \dfrac{d^2y}{dx^2}=e^x(\sin 5x+5\cos 5x)+e^x(5\cos 5x-25\sin 5x)
Product rule: \dfrac{dy}{dx}=6e^{6x}\cos 3x-3e^{6x}\sin 3x=e^{6x}(6\cos 3x-3\sin 3x)
Differentiate again: \dfrac{d^2y}{dx^2}=6e^{6x}(6\cos 3x-3\sin 3x)+e^{6x}(-18\sin 3x-9\cos 3x)
\dfrac{dy}{dx}=\dfrac{1}{1+x^2}
Differentiate using the chain rule on (1+x^2)^{-1}: \dfrac{d^2y}{dx^2}=-\dfrac{2x}{(1+x^2)^2}
\dfrac{dy}{dx}=\dfrac{1}{x\log x}
Differentiate (x\log x)^{-1} using the chain rule, with \dfrac{d}{dx}(x\log x)=\log x+1: \dfrac{d^2y}{dx^2}=-\dfrac{\log x+1}{(x\log x)^2}
\dfrac{dy}{dx}=\cos(\log x)\cdot\dfrac{1}{x}
Differentiate \dfrac{\cos(\log x)}{x} using the quotient rule: \dfrac{d^2y}{dx^2}=\dfrac{x\left[-\sin(\log x)\cdot\frac{1}{x}\right]-\cos(\log x)}{x^2}
\dfrac{dy}{dx}=-5\sin x-3\cos x
\dfrac{d^2y}{dx^2}=-5\cos x+3\sin x=-(5\cos x-3\sin x)=-y
Given: y=\cos^{-1}x …(1)
From (1): x=\cos y …(2)
Differentiate (1): \dfrac{dy}{dx}=\dfrac{-1}{\sqrt{1-x^2}}. Using (2), 1-x^2=1-\cos^2y=\sin^2y, so \sqrt{1-x^2}=\sin y (for y\in(0,\pi)): \dfrac{dy}{dx}=\dfrac{-1}{\sin y}=-\csc y
Differentiate again w.r.t. x (chain rule through y): \dfrac{d^2y}{dx^2}=\csc y\cot y\cdot\dfrac{dy}{dx}=\csc y\cot y\cdot(-\csc y)
\dfrac{dy}{dx}=\dfrac{-3\sin(\log x)+4\cos(\log x)}{x}\;\Rightarrow\;x\dfrac{dy}{dx}=-3\sin(\log x)+4\cos(\log x)
Differentiate again (product rule on the left, chain rule on the right): \dfrac{dy}{dx}+x\dfrac{d^2y}{dx^2}=-\dfrac{3\cos(\log x)+4\sin(\log x)}{x}=-\dfrac{y}{x}
Multiply through by x: x\dfrac{dy}{dx}+x^2\dfrac{d^2y}{dx^2}=-y
\dfrac{dy}{dx}=Ame^{mx}+Bne^{nx}
\dfrac{d^2y}{dx^2}=Am^2e^{mx}+Bn^2e^{nx}
\dfrac{d^2y}{dx^2}-(m+n)\dfrac{dy}{dx}=Am^2e^{mx}+Bn^2e^{nx}-(m+n)\big(Ame^{mx}+Bne^{nx}\big)=-mn\big(Ae^{mx}+Be^{nx}\big)=-mny
\dfrac{dy}{dx}=3500e^{7x}-4200e^{-7x}
\dfrac{d^2y}{dx^2}=24500e^{7x}+29400e^{-7x}=49\big(500e^{7x}+600e^{-7x}\big)
Given: e^y(x+1)=1 …(1)
Differentiate (1) implicitly (product rule): e^y\cdot y'(x+1)+e^y=0\;\Rightarrow\;e^y\big[y'(x+1)+1\big]=0
Since e^y\ne0: y'(x+1)=-1 …(2)
Differentiate (2) w.r.t. x (product rule on the left): y''(x+1)+y'=0\;\Rightarrow\;y''=\dfrac{-y'}{x+1}
From (2), \dfrac{1}{x+1}=-y', so y''=-y'\cdot(-y')=(y')^2
Given: y=(\tan^{-1}x)^2 …(1)
Differentiate (1) (chain rule): y'=2\tan^{-1}x\cdot\dfrac{1}{1+x^2}\;\Rightarrow\;(1+x^2)y'=2\tan^{-1}x …(2)
Differentiate (2) w.r.t. x (product rule on the left): 2xy'+(1+x^2)y''=\dfrac{2}{1+x^2}
Multiply throughout by (1+x^2): 2x(1+x^2)y'+(1+x^2)^2y''=2
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