If y= (6 x^3-3 x^2-9 x )^ 10 , find d y d x

Question

If $y=\left(6 x^3-3 x^2-9 x\right)^{10}$, find $\frac{d y}{d x}$

✓

Answer

Given $y=\left(6 x^3-3 x^2-9 x\right)^{10}$
Let $u=6 x^3-3 x^2-9 x$
Then $y=u^{10}$
$
\begin{aligned}
& \begin{aligned}
\therefore \frac{d y}{d u} & =\frac{d}{d u}\left(u^{10}\right)=10 u^9 \\
& =10\left(6 x^3-3 x^2-9 x\right)^9
\end{aligned} \\
& \text { and } \frac{d u}{d x}=\frac{d}{d x}\left(6 x^3-3 x^2-9 x\right)
\end{aligned}
$
$
\begin{aligned}
& =6 \frac{d}{d x}\left(x^3\right)-3 \frac{d}{d x}\left(x^2\right)-9 \frac{d}{d x}(x) \\
& =6 \times 3 x^2-3 \times 2 x-9 \times 1 \\
& =18 x^2-6 x-9 \\
\therefore \frac{d y}{d x}= & \frac{d y}{d u} \cdot \frac{d u}{d x} \\
& =10\left(6 x^2-3 x^2-9 x\right)^9 \cdot\left(18 x^2-6 x-9\right) .
\end{aligned}
$

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