What is the derivative of $\frac{4}{x + 3}$ $4 / (x+3)$ ?

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Answer 1 · 2016-11-01T14:49:10

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

This problem can be solved with the use of the chain rule...

$y = 4 {(x + 3)}^{- 1}$ $y=4(x+3)^(-1)$

$y = 4 u^{- 1}$ $y=4u^(-1)$

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

$u = x + 3$ $u=x+3$

$\frac{d u}{d x} = 1$ $(du)/(dx)=1$

$\frac{d y}{d u} \cdot \frac{d u}{d x} = - \frac{4}{{(x + 3)}^{2}} = \frac{d y}{d x}$ $(dy)/(du)*(du)/(dx)=-4/(x+3)^2=(dy)/(dx)$

What is the derivative of 4x+34 / (x+3)?