How do you find the derivative of y=( (2x^3)*(e^(sinx))*(2^(cosx)) ) / (tanx-3^x)?

Jul 16, 2018

$y ' = \frac{u ' v - u v '}{v} ^ 2$ where $u ' = 6 {x}^{2} {e}^{\sin} \left(x\right) {2}^{\cos} \left(x\right) + 2 {x}^{3} {e}^{\sin} \left(x\right) \cos \left(x\right) {2}^{\cos} \left(x\right) + {2}^{\cos} \left(x\right) \ln \left(2\right) \left(- \sin \left(x\right)\right)$
$v = \tan \left(x\right) - {3}^{x}$
$v ' = 1 + {\tan}^{2} \left(x\right) - {3}^{x} \ln \left(3\right)$

Explanation:

We Need the rule

$\left(u v w\right) ' = u ' v w + u v ' w + u v w '$

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{v} ^ 2$

$u ' = 6 {x}^{2} {e}^{\sin} \left(x\right) {2}^{\cos} \left(x\right) + 2 {x}^{3} {e}^{\sin} \left(x\right) \cos \left(x\right) {2}^{\cos} \left(x\right) + {2}^{\cos} \left(x\right) \ln \left(2\right) \cdot \left(- \sin \left(x\right)\right)$

$v ' = 1 + {\tan}^{2} \left(x\right) - {3}^{x} \ln \left(3\right)$
Note that

$\left(\tan \left(x\right)\right) ' = 1 + {\tan}^{2} \left(x\right)$