S steezin Apr 2010 25 0 May 11, 2010 #1 Let y = g(h(x)) where h(x) = x^2/x+2. If g'(9/5) = -2, find dy/dx when x = 3

skeeter MHF Helper Jun 2008 16,216 6,764 North Texas May 11, 2010 #2 steezin said: Let y = g(h(x)) where h(x) = x^2/x+2. If g'(9/5) = -2, find dy/dx when x = 3 Click to expand... chain rule ... \(\displaystyle \frac{dy}{dx} = g'[h(x)] \cdot h'(x) \) note that \(\displaystyle h(3) = \frac{9}{5}\) finish it. Reactions: steezin

steezin said: Let y = g(h(x)) where h(x) = x^2/x+2. If g'(9/5) = -2, find dy/dx when x = 3 Click to expand... chain rule ... \(\displaystyle \frac{dy}{dx} = g'[h(x)] \cdot h'(x) \) note that \(\displaystyle h(3) = \frac{9}{5}\) finish it.

S steezin Apr 2010 25 0 May 11, 2010 #3 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 16,216 6,764 North Texas May 11, 2010 #4 \(\displaystyle \frac{dy}{dx} = g'[h(x)] \cdot h'(x) \) \(\displaystyle \frac{dy}{dx} = g'\left(\frac{x^2}{x+2}\right) \cdot \frac{(x+2)(2x)-(x^2)(1)}{(x+2)^2} \) now evaluate for \(\displaystyle x = 3\) Reactions: steezin

\(\displaystyle \frac{dy}{dx} = g'[h(x)] \cdot h'(x) \) \(\displaystyle \frac{dy}{dx} = g'\left(\frac{x^2}{x+2}\right) \cdot \frac{(x+2)(2x)-(x^2)(1)}{(x+2)^2} \) now evaluate for \(\displaystyle x = 3\)

S steezin Apr 2010 25 0 May 11, 2010 #5 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?

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 May 11, 2010 #6 \(\displaystyle \frac{dy}{dx} = g'\left(\frac{x^2}{x+2}\right) \cdot \frac{(x+2)(2x)-(x^2)(1)}{(x+2)^2} \) 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 Click to expand... Reactions: steezin

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