How do you differentiate $y=ae^x+b/v+c/v^2$ ?

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Answer 1 · 2016-11-12T16:46:49

I will assume that we need to find $dy/dx$ and that $a$ , $b$ , and $c$ are constants and that $v$ is some unspecified function of $x$ .

$y=ae^x+bv^-1 + cv^-2$

$dy/dx = ae^x-bv^(-1-1) (dv)/dx -2c v^(-2-1) (dv)/dx$

$= ae^x-bv^-2 (dv)/dx -2c v^-3 (dv)/dx$

$= ae^x - b/v^2 (dv)/dx - (2c)/v^3 (dv)/dx$

How do you differentiate y=ae^x+b/v+c/v^2y=ae^x+b/v+c/v^2?