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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle f(x) = \sin^{-1}{\left(5 x \right)}\).
Using the chain rule gives \(\displaystyle f'(x) = \frac{df}{du}\frac{du}{dx}\) where \(\displaystyle f(u) = \sin^{-1}{\left(u \right)}\) and \(\displaystyle u = 5 x\) gives \(\displaystyle f'(x) = \frac{5}{\sqrt{1 - 25 x^{2}}}\)
\begin{question}Find the derivative of $f(x) = \sin^{-1}{\left(5 x \right)}$.
\soln{9cm}{Using the chain rule gives $f'(x) = \frac{df}{du}\frac{du}{dx}$ where $f(u) = \sin^{-1}{\left(u \right)}$ and $u = 5 x$ gives $f'(x) = \frac{5}{\sqrt{1 - 25 x^{2}}}$}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Find the derivative of <img class="equation_image" title=" \displaystyle f(x) = \sin^{-1}{\left(5 x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Csin%5E%7B-1%7D%7B%5Cleft%285%20x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle f(x) = \sin^{-1}{\left(5 x \right)} " data-equation-content=" \displaystyle f(x) = \sin^{-1}{\left(5 x \right)} " /> . </p> </p><p> <p>Using the chain rule gives <img class="equation_image" title=" \displaystyle f'(x) = \frac{df}{du}\frac{du}{dx} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7Bdf%7D%7Bdu%7D%5Cfrac%7Bdu%7D%7Bdx%7D%20" alt="LaTeX: \displaystyle f'(x) = \frac{df}{du}\frac{du}{dx} " data-equation-content=" \displaystyle f'(x) = \frac{df}{du}\frac{du}{dx} " /> where <img class="equation_image" title=" \displaystyle f(u) = \sin^{-1}{\left(u \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%28u%29%20%3D%20%5Csin%5E%7B-1%7D%7B%5Cleft%28u%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle f(u) = \sin^{-1}{\left(u \right)} " data-equation-content=" \displaystyle f(u) = \sin^{-1}{\left(u \right)} " /> and <img class="equation_image" title=" \displaystyle u = 5 x " src="/equation_images/%20%5Cdisplaystyle%20u%20%3D%205%20x%20" alt="LaTeX: \displaystyle u = 5 x " data-equation-content=" \displaystyle u = 5 x " /> gives <img class="equation_image" title=" \displaystyle f'(x) = \frac{5}{\sqrt{1 - 25 x^{2}}} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7B5%7D%7B%5Csqrt%7B1%20-%2025%20x%5E%7B2%7D%7D%7D%20" alt="LaTeX: \displaystyle f'(x) = \frac{5}{\sqrt{1 - 25 x^{2}}} " data-equation-content=" \displaystyle f'(x) = \frac{5}{\sqrt{1 - 25 x^{2}}} " /> </p> </p>