# How do you find the derivative of x^3*arctan(7x)?

Apr 23, 2015

Firstly let's differentiate this function using implicit and logarithmic differentiation:

$q = \ln \left(\arctan \left(7 x\right)\right)$

${e}^{q} = \arctan \left(7 x\right)$

$\tan \left({e}^{q}\right) = 7 x$

${e}^{q} {\sec}^{2} \left({e}^{q}\right) \cdot \frac{\mathrm{dq}}{\mathrm{dx}} = 7$

$\arctan \left(7 x\right) \cdot \left({\tan}^{2} \left({e}^{q}\right) + 1\right) \frac{\mathrm{dq}}{\mathrm{dx}} = 7$

$\arctan \left(7 x\right) \cdot \left(49 {x}^{2} + 1\right) \frac{\mathrm{dq}}{\mathrm{dx}} = 7$

(dq)/(dx)=7/(arctan(7x)*(49x^2+1)

Alright, knowing this we can now differentiate ${x}^{3} \cdot \arctan \left(7 x\right)$ using implicit differentiation and the result above...

$y = {x}^{3} \cdot \arctan \left(7 x\right)$

$\ln y = \ln \left({x}^{3} \cdot \arctan \left(7 x\right)\right)$

$\ln y = \ln \left({x}^{3}\right) + \ln \left(\arctan \left(7 x\right)\right)$

$\ln y = 3 \ln x + \ln \left(\arctan \left(7 x\right)\right)$

1/y*(dy)/(dx)=3/x+7/(arctan(7x)*(49x^2+1)

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left\{\frac{3}{x} + \frac{7}{\arctan \left(7 x\right) \cdot \left(49 {x}^{2} + 1\right)}\right\}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{3} \cdot \arctan \left(7 x\right) \left\{\frac{3}{x} + \frac{7}{\arctan \left(7 x\right) \cdot \left(49 {x}^{2} + 1\right)}\right\}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{3} \cdot \arctan \left(7 x\right)}{x} + \frac{7 {x}^{3} \cdot \arctan \left(7 x\right)}{\arctan \left(7 x\right) \cdot \left(49 {x}^{2} + 1\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} \cdot \arctan \left(7 x\right) + \frac{7 {x}^{3}}{49 {x}^{2} + 1}$

I can also give you an alternative way of finding this derivative, using the product rule...

$y = {x}^{3} \cdot \arctan \left(7 x\right) = u \cdot v$

$u = {x}^{3}$, therefore $\frac{\mathrm{du}}{\mathrm{dx}} = 3 {x}^{2}$

$v = \arctan \left(7 x\right)$

$\tan v = 7 x$

${\sec}^{2} v \cdot \frac{\mathrm{dv}}{\mathrm{dx}} = 7$

$\left({\tan}^{2} v + 1\right) \cdot \frac{\mathrm{dv}}{\mathrm{dx}} = 7$

$\left(49 {x}^{2} + 1\right) \cdot \frac{\mathrm{dv}}{\mathrm{dx}} = 7$

$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{7}{49 {x}^{2} + 1}$

This means that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{3} \cdot \frac{7}{49 {x}^{2} + 1} + \arctan \left(7 x\right) \cdot 3 {x}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{7 {x}^{3}}{49 {x}^{2} + 1} + 3 {x}^{2} \cdot \arctan \left(7 x\right)$