Question
Differentiate the following w.r.t. x :
$\left(x^3-2 x-1\right)^5$
✓
Answer
Method 1: Let $y=\left(x^3-2 x-1\right)^5$ Put $u=x^3-2 x-1$. Then $y=u^5$
$\therefore \frac{d y}{d u}=\frac{d}{d u}\left(u^5\right)=5 u^4$
$\begin{aligned} & =5\left(x^3-2 x-1\right)^4 \\ \text { and } \frac{d u}{d x} & =\frac{d}{d x}\left(x^3-2 x-1\right) \\ & =3 x^2-2 \times 1-0=3 x^2-2 \\ \therefore \frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x} \\ & =5\left(x^3-2 x-1\right)^4\left(3 x^2-2\right) \\ & =5\left(3 x^2-2\right)\left(x^3-2 x-1\right)^4 .\end{aligned}$
Method 2: Let $y=\left(x^3-2 x-1\right)^5$ Differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(x^3-2 x-1\right)^5 \\ & =5\left(x^3-2 x-1\right)^4 \times \frac{d}{d x}\left(x^3-2 x-1\right) \\ & =5\left(x^3-2 x-1\right)^4 \times\left(3 x^2-2 \times 1-0\right) \\ & =5\left(3 x^2-2\right)\left(x^3-2 x-1\right)^4\end{aligned}$
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