Question
Find $f^{\prime}(x)$ if $f(x)=\left(x^{2}+3 x+4\right)^{7}$
✓
Answer
$f(x)=\left(x^2+3 x+4\right)^7$
Taking, $\mathrm{u}=\mathrm{x}^2+3 \mathrm{x}+4$
$f(x)=u^7$
As per the chain rule,
$f^{\prime}(\mathrm{x})=\frac{d}{d u}\left(u^7\right) \times \frac{d}{d x}\left(\mathrm{x}^2+3 \mathrm{x}+4\right)$
$ =7 u^6 \times(2 \mathrm{x}+3)$
Putting, $u=x^2+3 x+4$
$f^{\prime}(x)=7\left(x^2+3 x+4\right)^6 \cdot(2 x+3)$
Hence, $f(x)=\left(x^2+3 x+4\right)^7$ then
$f^{\prime}(x)=7\left(x^2+3 x+4\right)^6 \cdot(2 x+3)$
Taking, $\mathrm{u}=\mathrm{x}^2+3 \mathrm{x}+4$
$f(x)=u^7$
As per the chain rule,
$f^{\prime}(\mathrm{x})=\frac{d}{d u}\left(u^7\right) \times \frac{d}{d x}\left(\mathrm{x}^2+3 \mathrm{x}+4\right)$
$ =7 u^6 \times(2 \mathrm{x}+3)$
Putting, $u=x^2+3 x+4$
$f^{\prime}(x)=7\left(x^2+3 x+4\right)^6 \cdot(2 x+3)$
Hence, $f(x)=\left(x^2+3 x+4\right)^7$ then
$f^{\prime}(x)=7\left(x^2+3 x+4\right)^6 \cdot(2 x+3)$
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