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Chapter 3: Problem 76

Calculate the derivative of the following functions. \(y=\sqrt[5]{f(x) g(x)},\) where \(f\) and \(g\) are differentiable at \(x\)

Short Answer

Expert verified
Question: Calculate the derivative of the function \(y = \sqrt[5]{f(x)g(x)}\) with respect to \(x\). Answer: The derivative of the given function is \(y' = \frac{1}{5}\sqrt[5]{\frac{1}{(f(x)g(x))^4}}(f'(x)g(x)+f(x)g'(x))\).

Step by step solution

01

Rewrite the function in exponential form

To make differentiation easier, we rewrite the function as: \(y=\sqrt[5]{f(x)g(x)} = (f(x)g(x))^{\frac{1}{5}}\)
02

Apply the chain rule

Now let's apply the chain rule to differentiate the function. The chain rule is: \((u(v(x)))'=u'(v(x))v'(x)\) Let \(u(v) = v^{\frac{1}{5}}\) and \(v(x) = f(x)g(x)\). So, our function is \(y = u(v(x))\). Differentiating \(u\) with respect to \(v\), we get: \(u'(v) = \frac{1}{5}v^{-\frac{4}{5}}\) Now, we can calculate the derivative of \(y\) with respect to \(x\) using the chain rule: \(y' = u'(v(x))v'(x) = \frac{1}{5}(f(x)g(x))^{-\frac{4}{5}}(f'(x)g(x)+f(x)g'(x))\)
03

Simplify the expression

Finally, let's simplify the expression: \(y' = \frac{1}{5}(f(x)g(x))^{-\frac{4}{5}}(f'(x)g(x)+f(x)g'(x))\) We can rewrite the expression using the root instead of the exponential form for a clearer presentation: \(y' = \frac{1}{5}\sqrt[5]{\frac{1}{(f(x)g(x))^4}}(f'(x)g(x)+f(x)g'(x))\) Thus, the derivative of the given function is: \(y' = \frac{1}{5}\sqrt[5]{\frac{1}{(f(x)g(x))^4}}(f'(x)g(x)+f(x)g'(x))\)

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