Question

Use logarithmic differentiation to find the derivative of the function $f(x) = \frac{ {x}^{ \frac{2}{5} } \sqrt{ {x}^{2} + 1} }{ \sqrt[5]{ {x}^{4} + 1 } }$

Answer 1

we are given the following function:

$f(x)=\frac{x^{\frac{2}{5}}\sqrt[]{x^2+1}}{\sqrt[5]{x^4+1}}$

we are asked to find the derivative of this function. To do that, we will take logarithms on both sides of the equation, like this:

$\ln f(x)=\ln \frac{x^{\frac{2}{5}}\sqrt[]{x^2+1}}{\sqrt[5]{x^4+1}}$

Now we will use the following property of logarithms:

$\ln \frac{a}{b}=\ln a-\ln b$

applying that property we get:

$\ln f(x)=\ln x^{\frac{2}{5}}\sqrt[]{x^2+1}-\ln \sqrt[5]{x^4+1}$

Now we will use the following property of logarithms:

$\ln ab=\ln a+\ln b$

applying that property we get:

$\ln f(x)=\ln x^{\frac{2}{5}}+\ln \sqrt[]{x^2+1}-\ln \sqrt[5]{x^4+1}$

Now we'll make use of the following property of logarithms:

$\ln a^b=b\ln a$

We'll rewrite the roots first as fractional exponents:

$\ln f(x)=\ln x^{\frac{2}{5}}+\ln (x^2+1)^{\frac{1}{2}}-\ln (x^4+1)^{\frac{1}{5}}$

Now we apply the property:

$\ln f(x)=\frac{2}{5}\ln x+\frac{1}{2}\ln (x^2+1)-\frac{1}{5}\ln (x^4+1)$

Now we differentiate on both sides of the equation, following the rules of logarithmic differentiation, that is:

$\frac{d}{dx}(\ln f(x))=\frac{f^{\prime}(x)}{f(x)}$

Applying that to the equation:

$\frac{f^{\prime}(x)}{f(x)}=\frac{2}{5}(\frac{1}{x})+\frac{1}{2}(\frac{2x}{x^2+1})-\frac{1}{2}(\frac{4x^3}{x^4+1})$

Simplifying:

$\frac{f^{\prime}(x)}{f(x)}=\frac{2}{5x}+\frac{x}{x^2+1}-\frac{2x^3}{x^4+1}$

Now we solve for f'(x), like this:

$f^{\prime}(x)=f(x)(\frac{2}{5x}+\frac{x}{x^2+1}-\frac{2x^3}{x^4+1})$

Now we substitute the value of f(x)

$f^{\prime}(x)=\frac{x^{\frac{2}{5}}\sqrt[]{x^2+1}}{\sqrt[5]{x^4+1}}(\frac{2}{5x}+\frac{x}{x^2+1}-\frac{2x^3}{x^4+1})$

And this is the derivative of the function.