# How do you find the derivative of [x+(x+sin^2x)^3]^4?

$\frac{d}{\mathrm{dx}} {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{4}$
$= 4 {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{3} \cdot \left[1 + 3 {\left(x + {\sin}^{2} x\right)}^{2} \cdot \left(1 + \sin 2 x\right)\right]$
Take note: $\sin 2 x = 2 \cdot \sin x \cdot \cos x$

#### Explanation:

From the given

${\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{4}$

$\frac{d}{\mathrm{dx}} {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{4} =$
$4 \cdot {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{4 - 1} \cdot \frac{d}{\mathrm{dx}} \left(x + {\left(x + {\sin}^{2} x\right)}^{3}\right)$

$= 4 \cdot {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{3} \cdot \left(1 + 3 \cdot {\left(x + {\sin}^{2} x\right)}^{3 - 1} \cdot \frac{d}{\mathrm{dx}} \left(x + {\sin}^{2} x\right)\right)$

$= 4 \cdot {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{3} \cdot \left(1 + 3 \cdot {\left(x + {\sin}^{2} x\right)}^{2} \cdot \left(1 + 2 \cdot \sin x \cdot \cos x\right)\right)$

$= 4 \cdot {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{3} \cdot \left(1 + 3 \cdot {\left(x + {\sin}^{2} x\right)}^{2} \cdot \left(1 + \sin 2 x\right)\right)$

God bless....I hope the explanation is useful.