If $f (x) = - e^{x}$ $f(x)= - e^x$ and $g (x) = \frac{1}{\sqrt{1 - x}}$ $g(x) =1 / sqrt(1-x$ , how do you differentiate $f (g (x))$ $f(g(x))$ using the chain rule?

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If $f (x) = - e^{x}$ $f(x)= - e^x$ and $g (x) = \frac{1}{\sqrt{1 - x}}$ $g(x) =1 / sqrt(1-x$ , how do you differentiate $f (g (x))$ $f(g(x))$ using the chain rule?

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Answer 1 · 2017-11-22T01:04:32

Substitute g in for x to find f(g) from f(x), and recall $\frac{d}{d x} f (g (x)) = \frac{d f}{d g} \frac{d g}{d x}$ $d/dx f(g(x))= (df)/(dg) (dg)/dx$ . See explanation.

Explanation:

Recall that the chain rule states that if $y = f (g (x)), \frac{d y}{d x} = \frac{d f}{d g} \cdot \frac{d g}{d x}$ $y=f(g(x)), dy/dx =(df)/(dg)*(dg)/(dx)$

We have $f (x) = - e^{x}, g (x) = \frac{1}{\sqrt{1 - x}}$ $f(x) =-e^x, g(x) = 1/sqrt(1-x)$

(This step is unnecessary, but in case the student is curious, to find f(g) you substitute g in for x in f(x), making $f (g) = - e^{g}$ $f(g) = -e^g$

We know that $\frac{d}{d u} k e^{u} = k e^{u}$ $d/(du) ke^u=ke^u$ , and $\frac{d}{d x} (\frac{1}{{(1 - x)}^{\frac{1}{2}}}) = \frac{1}{2} \cdot \frac{1}{{(1 - x)}^{\frac{3}{2}}} = \frac{1}{2 {(1 - x)}^{\frac{3}{2}}}$ $d/dx(1/(1-x)^(1/2)) = 1/2 * 1/(1-x)^(3/2) = 1/(2(1-x)^(3/2))$ .
Thus...

$\frac{d y}{d x} = \frac{d f}{d g} \frac{d g}{d x} = \frac{1}{2 {(1 - x)}^{\frac{3}{2}}} \cdot - e^{\frac{1}{\sqrt{1 - x}}} = \frac{- e^{\frac{1}{\sqrt{1 - x}}}}{2 {(1 - x)}^{\frac{3}{2}}}$ $dy/dx = (df)/(dg) (dg)/(dx)= 1/(2(1-x)^(3/2)) *- e^(1/sqrt(1-x)) = (-e^(1/sqrt(1-x)))/(2(1-x)^(3/2))$

Answer 2 · 2017-11-22T01:17:12

$\frac{d}{d x} f (g (x)) = - \frac{e^{\sqrt{1 - x}}}{2 {(1 - x)}^{\frac{3}{2}}}$ $d/dx f(g(x)) = - (e^(sqrt(1-x)))/(2(1-x)^(3/2))$

Explanation:

$f (x) = - e^{x} and g (x) = \frac{1}{\sqrt{1 - x}}$ $f(x) = -e^x and g(x) = 1/sqrt(1-x)$

Replace $x$ $x$ in $f (x)$ $f(x)$ with g(x)#

$\to f (g (x)) = - e^{\frac{1}{\sqrt{i - x}}}$ $-> f(g(x)) = -e^(1/sqrt(i-x))$

$= - e^{{(1 - x)}^{- \frac{1}{2}}}$ $= -e^((1-x)^(-1/2))$

Apply the chain rule and standard differential

$\frac{d}{d x} f (g (x)) = - e^{{(1 - x)}^{- \frac{1}{2}}} \cdot \frac{d}{d x} {(1 - x)}^{- \frac{1}{2}}$ $d/dx f(g(x)) = -e^((1-x)^(-1/2)) * d/dx (1-x)^(-1/2)$

Apply the chain rule and power rule

$\frac{d}{d x} f (g (x)) = - e^{{(1 - x)}^{- \frac{1}{2}}} \cdot (- \frac{1}{2}) {(1 - x)}^{- \frac{3}{2}} \cdot \frac{d}{d x} (1 - x)$ $d/dx f(g(x)) = -e^((1-x)^(-1/2)) * (-1/2)(1-x)^(-3/2)* d/dx (1-x)$

$= - e^{{(1 - x)}^{- \frac{1}{2}}} \cdot (- \frac{1}{2}) {(1 - x)}^{- \frac{3}{2}} \cdot (- 1)$ $= -e^((1-x)^(-1/2)) * (-1/2)(1-x)^(-3/2)* (-1)$

$= - e^{\frac{1}{\sqrt{1 - x}}} \cdot \frac{1}{2 {(1 - x)}^{\frac{3}{2}}}$ $= -e^(1/sqrt(1-x)) * 1/(2(1-x)^(3/2))$

$= - \frac{e^{\sqrt{1 - x}}}{2 {(1 - x)}^{\frac{3}{2}}}$ $= - (e^(sqrt(1-x)))/(2(1-x)^(3/2))$

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If $f (x) = - e^{x}$ $f(x)= - e^x$ and $g (x) = \frac{1}{\sqrt{1 - x}}$ $g(x) =1 / sqrt(1-x$ , how do you differentiate $f (g (x))$ $f(g(x))$ using the chain rule?

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If f(x)=−exf(x)= - e^x and g(x)=1√1−xg(x) =1 / sqrt(1-x , how do you differentiate f(g(x))f(g(x)) using the chain rule?

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If $f (x) = - e^{x}$ $f(x)= - e^x$ and $g (x) = \frac{1}{\sqrt{1 - x}}$ $g(x) =1 / sqrt(1-x$ , how do you differentiate $f (g (x))$ $f(g(x))$ using the chain rule?