derivatives

Apr 2010
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Let y = g(h(x)) where h(x) = x^2/x+2. If g'(9/5) = -2, find dy/dx when x = 3
 

skeeter

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Jun 2008
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North Texas
Let y = g(h(x)) where h(x) = x^2/x+2. If g'(9/5) = -2, find dy/dx when x = 3
chain rule ...

\(\displaystyle
\frac{dy}{dx} = g'[h(x)] \cdot h'(x)
\)

note that \(\displaystyle h(3) = \frac{9}{5}\)

finish it.
 
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Apr 2010
25
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I know how to derive the inner function but I don't know how to find the derivative of g(h(x))?.
 

skeeter

MHF Helper
Jun 2008
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North Texas
\(\displaystyle
\frac{dy}{dx} = g'[h(x)] \cdot h'(x)
\)

\(\displaystyle
\frac{dy}{dx} = g'\left(\frac{x^2}{x+2}\right) \cdot \frac{(x+2)(2x)-(x^2)(1)}{(x+2)^2}
\)

now evaluate for \(\displaystyle x = 3\)
 
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Apr 2010
25
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So I would get dy/dx = x^4 + 4x^3/(x + 2)^3? and sub in 3 for x. The g' is confusing me, does it just disappear or something?
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
\(\displaystyle
\frac{dy}{dx} = g'\left(\frac{x^2}{x+2}\right) \cdot \frac{(x+2)(2x)-(x^2)(1)}{(x+2)^2}
\)

sub in 3 ...

\(\displaystyle \frac{dy}{dx} = \textcolor{red}{g'\left(\frac{3^2}{3+2}\right)} \cdot \frac{(3+2)(2\cdot 3)-(3^2)(1)}{(3+2)^2}\)


now, from your original post ...

If g'(9/5) = -2, find dy/dx when x = 3
 
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