How do you find the derivative of #((2x+1)^(5)/(x^(2)+1)^(1/2))#?

Question

Emilia Aguilar · Accepted Answer

#:.f'(x)=(((2x+1)^5)/((x^2+1)^(1/2)))[10/(2x+1)-x/(x^2+1)]# Let #f(x)=(((2x+1)^5)/((x^2+1)^(1/2)))# #:. lnf (x)=5ln (2x+1)-1/2ln (x^2+1)# Now differentiate with respect to #x# #(f'(x))/(f (x))=10/(2x+1)-x/(x^2+1)# #:.f'(x)=(((2x+1)^5)/((x^2+1)^(1/2)))[10/(2x+1)-x/(x^2+1)]#

Ezekiel Alexander · Answer

undefined To find the derivative of $\frac{{(2x+1)^5}}{{(x^2+1)^{1/2}}}$, you can use the quotient rule. The quotient rule states that if you have a function in the form $f(x) = \frac{{g(x)}}{{h(x)}}$, then its derivative is given by $$f'(x) = \frac{{g'(x)h(x) - g(x)h'(x)}}{{[h(x)]^2}}.$$

Here, $g(x) = (2x+1)^5$ and $h(x) = (x^2+1)^{1/2}$.

We'll need to find the derivatives of $g(x)$ and $h(x)$ first using the chain rule and power rule, respectively.

$g'(x) = 5(2x+1)^4(2)$  
$h'(x) = \frac{{1}}{{2}}(x^2+1)^{-\frac{{1}}{{2}}}(2x)$

Now, apply the quotient rule:

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

Simplify the expression:

$$f'(x) = \frac{{10x(2x+1)^4(x^2+1)^{1/2} - (2x+1)^5(x^2+1)^{-\frac{{1}}{{2}}}}}{{(x^2+1)}}$$

That's the derivative of the given function.