Math  /  Calculus

QuestionFind yy^{\prime}. y=(x+1x7)7y=\begin{array}{l} y=\left(\frac{x+1}{x-7}\right)^{7} \\ y^{\prime}=\square \end{array}

Studdy Solution
Apply the chain rule to find y y' .
The chain rule states that dydx=dydududx \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} .
y=7u68(x7)2 y' = 7u^6 \cdot \frac{-8}{(x-7)^2}
Substitute back u=x+1x7 u = \frac{x+1}{x-7} :
y=7(x+1x7)68(x7)2 y' = 7\left(\frac{x+1}{x-7}\right)^6 \cdot \frac{-8}{(x-7)^2}
y=56(x+1x7)61(x7)2 y' = -56 \left(\frac{x+1}{x-7}\right)^6 \cdot \frac{1}{(x-7)^2}
The derivative y y' is:
y=56(x+1)6(x7)8 y' = \frac{-56(x+1)^6}{(x-7)^8}

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