Find f '(a). f(x) = √5x + 1 I Know i can turn it into (5x+1)^.5 but im lost from here on out. __________________________________________________ ___
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Originally Posted by Evan.Kimia Find f '(a). f(x) = √5x + 1 I Know i can turn it into (5x+1)^.5 but im lost from here on out. __________________________________________________ ___ is it $\displaystyle f(x)=\sqrt {5x}+1 \quad or \quad f(x)=\sqrt {5} {x}+1$ ?
the square root covers that whole function.
$\displaystyle f(x)=\sqrt{(5x+1)}$ use chain rule $\displaystyle f'(x)=\frac{d}{dx}f(x)=\frac{d{f(x)}}{dt} \cdot \frac{dt}{dx} $ assume t=5 x+1 so that f(x)= $\displaystyle t^{\frac{1}{2}} $ if stuck see spoiler Spoiler: $\displaystyle \frac{d{f(x)}}{dt} = \frac{1}{2} t^{\frac{-1}{2}}=\frac{1}{2 \sqrt {t}}=\frac{1}{2 \sqrt {5 x+1 }}$ $\displaystyle \frac{dt}{dx}=\frac{d{(5 x+1) }}{dx}=5$ $\displaystyle f'(x)=\frac{d{f(x)}}{dt} \cdot \frac{dt}{dx} =\frac{1}{2 \sqrt {5 x+1 }} \cdot 5=\frac{5}{2 \sqrt {5 x+1 }}$
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