# How do you differentiate  y = 4(x-(3-5x)^4)^2 using the chain rule?

$\frac{\mathrm{dy}}{\mathrm{dx}} = 8 x + 160 x {\left(3 - 5 x\right)}^{3} - 8 {\left(3 - 5 x\right)}^{4} - 160 {\left(3 - 5 x\right)}^{7}$

#### Explanation:

let us start from the given equation $y = 4 {\left(x - {\left(3 - 5 x\right)}^{4}\right)}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} 4 {\left(x - {\left(3 - 5 x\right)}^{4}\right)}^{2} = 4 \cdot \frac{d}{\mathrm{dx}} {\left(x - {\left(3 - 5 x\right)}^{4}\right)}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 \cdot 2 \cdot {\left(x - {\left(3 - 5 x\right)}^{4}\right)}^{2 - 1} \cdot \frac{d}{\mathrm{dx}} \left(x - {\left(3 - 5 x\right)}^{4}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 8 \cdot {\left(x - {\left(3 - 5 x\right)}^{4}\right)}^{1} \cdot \left[\frac{d}{\mathrm{dx}} \left(x\right) - \frac{d}{\mathrm{dx}} {\left(3 - 5 x\right)}^{4}\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 8 \cdot \left(x - {\left(3 - 5 x\right)}^{4}\right) \cdot \left[1 - 4 {\left(3 - 5 x\right)}^{4 - 1} \cdot \frac{d}{\mathrm{dx}} \left(3 - 5 x\right)\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 8 \cdot \left(x - {\left(3 - 5 x\right)}^{4}\right) \cdot \left[1 - 4 {\left(3 - 5 x\right)}^{3} \left(0 - 5\right)\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 8 \cdot \left(x - {\left(3 - 5 x\right)}^{4}\right) \cdot \left[1 + 20 {\left(3 - 5 x\right)}^{3}\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 8 \left(x - {\left(3 - 5 x\right)}^{4} + 20 x {\left(3 - 5 x\right)}^{3} - 20 {\left(3 - 5 x\right)}^{7}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 8 x + 160 x {\left(3 - 5 x\right)}^{3} - 8 {\left(3 - 5 x\right)}^{4} - 160 {\left(3 - 5 x\right)}^{7}$

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