How do you use the chain rule to differentiate $y = 2 {(x + 3)}^{\frac{1}{2}}$ $y=2(x+3)^(1/2)$ ?

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Answer 1 · 2018-06-05T18:04:07

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

$y (x) = 2 {(x + 3)}^{\frac{1}{2}}$ $y(x)=2(x+3)^(1/2$

Let $u (x) = x + 3$ $u(x)=x+3$

Differentiating this with powers rule:

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

This means $y (u) = 2 u^{\frac{1}{2}}$ $y(u)=2u^(1/2)$

Differentiate this with the powers rule.

$\frac{d y}{d u} = u^{- \frac{1}{2}}$ $dy/(du)=u^(-1/2$

Now, we can use the fact that

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $dy/dx=dy/(du) *(du)/dx$

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

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

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

How do you use the chain rule to differentiate y=2(x+3)12y=2(x+3)^(1/2)?