How do you differentiate #g(t)=1/(t^4+1)^3#?

Answer 1

To differentiate this function, you need to use the chain rule. When you do so, you get #g'(t)=(-12t^3)/(t^4+1)^4# .

Whenever you see a function nested inside another function—a composition of functions—and you want to find the derivative of the whole thing, you need to use the chain rule. In symbols, the chain rule states that #f@h=f'(h)*h'#. In this case, since the function #t^4+1# is then cubed and reciprocated, it is a composition of functions and the chain rule needs to be applied. #f(x)=x^-3# and #h(x)=t^4+1#, so #g'(x)=f'(t^4+1)*h'=-3(t^4+1)^-2*4t^3=(-12t^3)/(t^4+1)^4#.

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

Answer 2

Luke Alvarez

To differentiate ( g(t) = \frac{1}{(t^4 + 1)^3} ), you can use the chain rule. First, rewrite the function as ( g(t) = (t^4 + 1)^{-3} ). Then, differentiate using the chain rule:

[ g'(t) = -3(t^4 + 1)^{-4} \cdot 4t^3 ]

Simplify:

[ g'(t) = -12t^3(t^4 + 1)^{-4} ]

So, the derivative of ( g(t) ) with respect to ( t ) is ( -12t^3(t^4 + 1)^{-4} ).

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.