How do you find the derivative of $y = (\frac{2}{x^{2}}) - 5 x^{4}$ $y=(2/(x^2))-5x^4$ ?

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Answer 1 · 2018-05-21T04:58:13

$\frac{d y}{d x} = - \frac{4}{x^{3}} - 20 x^{3}$ $(dy)/(dx)=-4/x^3-20x^3$

We can use here the derivative of $x^{n}$ $x^n$ , which is $n x^{n - 1}$ $nx^(n-1)$

Hence as $y = (\frac{2}{x^{2}}) - 5 x^{4} = 2 x^{- 2} - 5 x^{4}$ $y=(2/x^2)-5x^4=2x^(-2)-5x^4$

$\frac{d y}{d x} = 2 \cdot (- 2) \cdot x^{- 2 - 1} - 5 \cdot 4 x^{4 - 1}$ $(dy)/(dx)=2*(-2)*x^(-2-1)-5*4x^(4-1)$

= $- 4 x^{- 3} - 20 x^{3}$ $-4x^(-3)-20x^3$

= $- \frac{4}{x^{3}} - 20 x^{3}$ $-4/x^3-20x^3$

How do you find the derivative of y=(2x2)−5x4y=(2/(x^2))-5x^4?