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

Answer 1

#f'(g(x))=(5sec^2(sqrt(5x-1))tan(sqrt(5x-1)))/(sqrt(5x-1))#

To find #f'(g(x))#, we can start by finding #f(g(x))# and then take the derivative. We substitute #sqrt(5x-1)# into #f(x)#, so we have:

#f(g(x))=f(sqrt(5x-1))=tan^2(sqrt(5x-1))#

Now we take the derivative using the chain rule.

#2tan(sqrt(5x-1))*sec^2(sqrt(5x-1))*1/2(5x-1)^(-1/2)*5#

#=>(5sec^2(sqrt(5x-1))tan(sqrt(5x-1)))/(sqrt(5x-1))#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Charles Anctil

To find ( f'(g(x)) ), you need to use the chain rule, which states that ( (f(g(x)))' = f'(g(x)) \cdot g'(x) ).

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

We first find ( f'(x) ) and ( g'(x) ):

[ f'(x) = 2\tan(x) \cdot \sec^2(x) ] [ g'(x) = \frac{1}{2\sqrt{5x - 1}} \cdot 5 = \frac{5}{2\sqrt{5x - 1}} ]

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

[ f'(g(x)) = 2\tan(\sqrt{5x - 1}) \cdot \sec^2(\sqrt{5x - 1}) \cdot \frac{5}{2\sqrt{5x - 1}} ]

Thus, ( f'(g(x)) = 5\tan(\sqrt{5x - 1}) \cdot \sec^2(\sqrt{5x - 1}) \cdot \frac{1}{\sqrt{5x - 1}} ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.