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

Question

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

Samantha Adkison · Accepted Answer

#f'(g (x))=tan ((sqrt (5x-1))/2)xxsec^2 ((sqrt (5x-1))/2)# #f'(x)=??# #" "# #f (x) " "# is a composite function composed of two functions #" "# Trigonometric function #u (x)=sec(x/2)# and #" "# Polynomial function #v (x)=x^2# #" "# #f'(x)# is determined by applying the chain rule . #" "# #f (x)=v (u (x))# #" "# #f'(x)=v'(u (x)) xx u'(x)# #" "# #v'(x)=(x^2)'=2x# #" "# #u'(x)=(sec(x/2))'=(x/2)'xxsec'(x/2)# #" "# #u'(x)=1/2xxtan(x/2)sec (x/2)# #" "# So,#f'(x)# is determined by substituting the differentiated functions #" "# #f'(x)=v'(u (x)) xx u'(x)=2 (u (x))xx1/2xxtan(x/2)sec (x/2)# #" "# #f'(x)=2sec (x/2)xx1/2xxtan(x/2)sec (x/2)# #" "# #f'(x)=tan (x/2)xxsec^2 (x/2)# #" "# #f'(g (x))=tan ((g (x))/2)xxsec^2 ((g (x))/2)# #" "# #f'(g (x))=tan ((sqrt (5x-1))/2)xxsec^2 ((sqrt (5x-1))/2)#

Ezra Adams · Answer

undefined To find $ f'(g(x)) $, first find $ g'(x) $ and then substitute into $ f'(x) $:

$$ g'(x) = \frac{d}{dx}\left(\sqrt{5x - 1}\right) = \frac{1}{2\sqrt{5x - 1}} \cdot 5 = \frac{5}{2\sqrt{5x - 1}} $$

Now, substitute $ g(x) $ into $ f'(x) $:

$$ f'(g(x)) = f'\left(\sqrt{5x - 1}\right) = \frac{d}{dx}\left(\sec^2\left(\frac{\sqrt{5x - 1}}{2}\right)\right) = \frac{d}{du}\left(\sec^2\left(\frac{u}{2}\right)\right) \cdot \frac{d}{dx}\left(\sqrt{5x - 1}\right) $$

Where $ u = \sqrt{5x - 1} $.

Using the chain rule and the derivative of $ \sec^2(u/2) $:

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

Substituting back $ u = \sqrt{5x - 1} $:

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