Question

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

Answer 1

`y'=(u'v-uv')/v^2` where `u'=6x^2e^sin(x)2^cos(x)+2x^3e^sin(x)cos(x)2^cos(x)+2^cos(x)ln(2)(-sin(x))`
`v=tan(x)-3^x`
`v'=1+tan^2(x)-3^xln(3)`

Explanation:

We Need the rule

`(uvw)'=u'vw+uv'w+uvw'`

`(u/v)'=(u'v-uv')/v^2`

`u'=6x^2e^sin(x)2^cos(x)+2x^3e^sin(x)cos(x)2^cos(x)+2^cos(x)ln(2)*(-sin(x))`

`v'=1+tan^2(x)-3^xln(3)`
Note that

`(tan(x))'=1+tan^2(x)`