How do you find the derivative of $f (y) = 3 y^{- 1} + 4 y$ $f(y)=3y^-1+4y$ ?

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Answer 1 · 2017-01-08T18:03:44

$\frac{d}{d y} (3 y^{- 1} + 4 y) = - 3 y^{- 2} + 4$ $d/(dy) (3y^(-1)+4y) =-3y^(-2)+4$

Using the general rule:

$\frac{d}{d y} (y^{n}) = n y^{n - 1}$ $d/(dy) (y^n) = ny^(n-1)$

And as the differentiation is linear, so:

$\frac{d}{d y} (a f (y) + b g (y)) = a (\frac{d}{d y} f (y)) + b (\frac{d}{d y} g (y))$ $d/(dy) (af(y)+bg(y))= a (d/(dy) f(y) )+ b(d/(dy) g(y))$

we have:

$\frac{d}{d y} (3 y^{- 1} + 4 y) = - 3 y^{- 2} + 4$ $d/(dy) (3y^(-1)+4y) =-3y^(-2)+4$

How do you find the derivative of f(y)=3y−1+4yf(y)=3y^-1+4y?