Question
Diffrentiate the following w.r.t.x
$\frac{\left(x^2+2\right)^4}{\sqrt{x^2+5}}$
✓
Answer
Let $y=\frac{\left(x^2+2\right)^4}{\sqrt{x^2+5}}$
Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left[\frac{\left(x^2+2\right)^4}{\sqrt{x^2+5}}\right]$
$\begin{aligned} & =\frac{\sqrt{x^2+5} \cdot \frac{d}{d x}\left(x^2+2\right)^4-\left(x^2+2\right)^4 \cdot \frac{d}{d x}\left(\sqrt{x^2+5}\right)}{\left(\sqrt{x^2+5}\right)^2} \\ & \sqrt{x^2+5} \times 4\left(x^2+2\right)^3 \cdot \frac{d}{d x}\left(x^2+2\right)- \\ & =\frac{\left(x^2+2\right)^4 \times \frac{1}{2 \sqrt{x^2+5}} \cdot \frac{d}{d x}\left(x^2+5\right)}{x^2+5} \\ & =\frac{\sqrt{x^2+5} \times 4\left(x^2+2\right)^3 \cdot(2 x+0)-\frac{\left(x^2+2\right)^4}{2 \sqrt{x^2+5}} \times(2 x+0)}{x^2+5} \\ & =\frac{8 x\left(x^2+5\right)\left(x^2+2\right)^3-x\left(x^2+2\right)^4}{\left(x^2+5\right)^{\frac{3}{2}}} \\ & =\frac{x\left(x^2+2\right)^3\left[8\left(x^2+5\right)-\left(x^2+2\right)\right]}{\left(x^2+5\right)^{\frac{3}{2}}} \\ & =\frac{x\left(x^2+2\right)^3\left(8 x^2+40-x^2-2\right)}{\left(x^2+5\right)^{\frac{3}{2}}} \\ & =\frac{x\left(x^2+2\right)^3\left(7 x^2+38\right)}{\left(x^2+5\right)^{\frac{3}{2}}} . \\ & \end{aligned}$
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