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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation2 Marks
Question
Differentiate the following w. r. t. x.$y=\left(x^3+2 x-3\right)^4(x+\cos x)^3$
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Answer

$ y=\left(x^3+2 x-3\right)^4(x+\cos x)^3 $ Differentiate $w . r . t . x$ $ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\left(x^3+2 x-3\right)^4(x+\cos x)^3\right] \\ & =\left(x^3+2 x-3\right)^4 \cdot \frac{d}{d x}(x+\cos x)^3+(x+\cos x)^3 \cdot \frac{d}{d x}\left(x^3+2 x-3\right)^4 \end{aligned} $$\begin{aligned} & =\left(x^3+2 x-3\right)^4 \cdot 3(x+\cos x)^2 \cdot \frac{d}{d x}(x+\cos x)+(x+\cos x)^3 \cdot 4\left(x^3+2 x-3\right)^3 \cdot \frac{d}{d x}\left(x^3+2 x-3\right) \\ & =\left(x^3+2 x-3\right)^4 \cdot 3(x+\cos x)^2(1-\sin x)+(x+\cos x)^3 \cdot 4\left(x^3+2 x-3\right)^3\left(3 x^2+2\right) \\ \therefore \quad \frac{d y}{d x} & =3\left(x^3+2 x-3\right)^4(x+\cos x)^2(1-\sin x)+4\left(3 x^2+2\right)\left(x^3+2 x-3\right)^3(x+\cos x)^3\end{aligned}$

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