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Exercise 8.2 · Q1

Q.Find differential dydy for each of the following functions:

(i) y=(1−2x)33−4xy=\dfrac{(1-2x)^3}{3-4x}
(ii) y=(3+sin⁡(2x))2/3y=(3+\sin(2x))^{2/3}
(iii) y=ex2−5x+7cos⁡(x2−1)y=e^{x^2-5x+7}\cos(x^2-1)
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✓ Free question

Each differential is dy=y′(x) dxdy=y'(x)\,dx; compute y′y' by the product/quotient/chain rules as appropriate.

Part (i): y=(1−2x)33−4xy=\dfrac{(1-2x)^3}{3-4x}. Write y=uvy=uv with u=(1−2x)3, v=(3−4x)−1u=(1-2x)^3,\ v=(3-4x)^{-1}.

u′=3(1−2x)2(−2)=−6(1−2x)2u'=3(1-2x)^2(-2)=-6(1-2x)^2. v′=−(3−4x)−2(−4)=4(3−4x)2\quad v'=-(3-4x)^{-2}(-4)=\dfrac{4}{(3-4x)^2}.

By the product rule, y′=u′v+uv′=−6(1−2x)23−4x+4(1−2x)3(3−4x)2=−6(1−2x)2(3−4x)+4(1−2x)3(3−4x)2y'=u'v+uv'=\dfrac{-6(1-2x)^2}{3-4x}+\dfrac{4(1-2x)^3}{(3-4x)^2}=\dfrac{-6(1-2x)^2(3-4x)+4(1-2x)^3}{(3-4x)^2}.

Factor (1−2x)2(1-2x)^2 from the numerator: −6(3−4x)+4(1−2x)=−18+24x+4−8x=16x−14=2(8x−7)-6(3-4x)+4(1-2x)=-18+24x+4-8x=16x-14=2(8x-7).

So y′=2(8x−7)(1−2x)2(3−4x)2y'=\dfrac{2(8x-7)(1-2x)^2}{(3-4x)^2}, hence dy=2(8x−7)(1−2x)2(3−4x)2 dxdy=\dfrac{2(8x-7)(1-2x)^2}{(3-4x)^2}\,dx.

Part (ii): y=(3+sin⁡2x)2/3y=(3+\sin2x)^{2/3}. By the chain rule, y′=23(3+sin⁡2x)−1/3⋅(2cos⁡2x)=4cos⁡2x3(3+sin⁡2x)1/3y'=\dfrac23(3+\sin2x)^{-1/3}\cdot(2\cos2x)=\dfrac{4\cos2x}{3(3+\sin2x)^{1/3}}.

So dy=4cos⁡2x3(3+sin⁡2x)1/3 dxdy=\dfrac{4\cos2x}{3(3+\sin2x)^{1/3}}\,dx.

Part (iii): y=ex2−5x+7cos⁡(x2−1)y=e^{x^2-5x+7}\cos(x^2-1). Product rule with u=ex2−5x+7, v=cos⁡(x2−1)u=e^{x^2-5x+7},\,v=\cos(x^2-1):

u′=ex2−5x+7(2x−5)u'=e^{x^2-5x+7}(2x-5). v′=−sin⁡(x2−1)(2x)\quad v'=-\sin(x^2-1)(2x).

y′=u′v+uv′=ex2−5x+7(2x−5)cos⁡(x2−1)−2x ex2−5x+7sin⁡(x2−1)=ex2−5x+7[(2x−5)cos⁡(x2−1)−2xsin⁡(x2−1)]y'=u'v+uv'=e^{x^2-5x+7}(2x-5)\cos(x^2-1)-2x\,e^{x^2-5x+7}\sin(x^2-1)=e^{x^2-5x+7}\big[(2x-5)\cos(x^2-1)-2x\sin(x^2-1)\big].

So dy=ex2−5x+7[(2x−5)cos⁡(x2−1)−2xsin⁡(x2−1)]dxdy=e^{x^2-5x+7}\big[(2x-5)\cos(x^2-1)-2x\sin(x^2-1)\big]dx.

✓Final answer

(i) dy=2(8x−7)(1−2x)2(3−4x)2 dxdy=\dfrac{2(8x-7)(1-2x)^2}{(3-4x)^2}\,dx (ii) dy=4cos⁡2x3(3+sin⁡2x)1/3 dxdy=\dfrac{4\cos2x}{3(3+\sin2x)^{1/3}}\,dx (iii) dy=ex2−5x+7[(2x−5)cos⁡(x2−1)−2xsin⁡(x2−1)]dxdy=e^{x^2-5x+7}\big[(2x-5)\cos(x^2-1)-2x\sin(x^2-1)\big]dx

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