How do you find the derivative of $y = \frac{(2 x^{3}) \cdot (e^{sin x}) \cdot (2^{cos x})}{tan x - 3^{x}}$ $y=( (2x^3)(e^(sinx))(2^(cosx)) ) / (tanx-3^x)$ ?

Question

1814 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-16T15:30:11

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

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

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

How do you find the derivative of y=( (2x^3)*(e^(sinx))*(2^(cosx)) ) / (tanx-3^x)y=(2x3)⋅(esinx)⋅(2cosx)tanx−3xy=( (2x^3)*(e^(sinx))*(2^(cosx)) ) / (tanx-3^x)?