If #f(x) =tan^2(x/2) # and #g(x) = sqrt(5x-1 #, what is #f'(g(x)) #?

Question

If #f(x) =tan^2(x/2) # and #g(x) = sqrt(5x-1  #, what is #f'(g(x)) #?

Christopher Agresta · Accepted Answer

#f'(g(x))=tan(sqrt(5x-1)/2)xxsec^2(sqrt(5x-1)/2)# As #f(x)=tan^2(x/2)#, #f'(x)=(d(tan^2(x/2)))/(d(tanx/2))##xx##(d(tanx/2))/(d(x/2))##xx##(d(x/2))/dx# Hence #f'(x)=2tan(x/2)xxsec^2(x/2)xx1/2# or #f'(x)=tan(x/2)xxsec^2(x/2)# Hence #f'(g(x))=tan(g(x)/2)xxsec^2(g(x)/2)# and as #g(x)=sqrt(5x-1)# #f'(g(x))=tan(sqrt(5x-1)/2)xxsec^2(sqrt(5x-1)/2)#

Samantha Adkison · Answer

undefined To find $ f'(g(x)) $, we first need to find $ f'(x) $ and $ g'(x) $, and then substitute $ g(x) $ into $ f'(x) $.

Given $ f(x) = \tan^2(x/2) $ and $ g(x) = \sqrt{5x - 1} $:

1. Find $ f'(x) $:
   $ f(x) = \tan^2(x/2) $
   $ f'(x) = \frac{d}{dx}[\tan^2(x/2)] $
   $ f'(x) = 2\tan(x/2) \sec^2(x/2) $

2. Find $ g'(x) $:
   $ g(x) = \sqrt{5x - 1} $
   $ g'(x) = \frac{d}{dx}[\sqrt{5x - 1}] $
   $ g'(x) = \frac{1}{2\sqrt{5x - 1}} \cdot 5 $
   $ g'(x) = \frac{5}{2\sqrt{5x - 1}} $

3. Substitute $ g(x) $ into $ f'(x) $:
   $ f'(g(x)) = 2\tan(g(x)/2) \sec^2(g(x)/2) $
   $ f'(g(x)) = 2\tan\left(\frac{\sqrt{5x - 1}}{2}\right) \sec^2\left(\frac{\sqrt{5x - 1}}{2}\right) $

Therefore, $ f'(g(x)) = 2\tan\left(\frac{\sqrt{5x - 1}}{2}\right) \sec^2\left(\frac{\sqrt{5x - 1}}{2}\right) $.

Luke Alvarez · Answer

undefined $$ f'(g(x)) = \frac{f'(u)}{g'(x)} $$

Where $ u = g(x) $

$$ f(u) = \tan^2\left(\frac{u}{2}\right) $$
$$ f'(u) = 2\tan\left(\frac{u}{2}\right)\sec^2\left(\frac{u}{2}\right) $$

$$ g(x) = \sqrt{5x - 1} $$
$$ g'(x) = \frac{5}{2\sqrt{5x - 1}} $$

$$ f'(g(x)) = \frac{2\tan\left(\frac{\sqrt{5x - 1}}{2}\right)\sec^2\left(\frac{\sqrt{5x - 1}}{2}\right)}{\frac{5}{2\sqrt{5x - 1}}} $$

$$ f'(g(x)) = \frac{4\tan\left(\frac{\sqrt{5x - 1}}{2}\right)\sec^2\left(\frac{\sqrt{5x - 1}}{2}\right)}{5} $$