Given the following functions: f(u)=tan(u) and g(x)=x^3. Find:
f ′(g(x))=
(f∘g)′(x)=
ps; there's a differentiation sign there
This is either a homework dump or a test or just for fun.
All of those problems have been addressed in earlier "calculus" posts, anyway.
Why are you switching names?
Jeff Bezos
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Identity complex?
I'm switching names for fun but I really need help. It was addressed but the answer is wrong
To find f ′(g(x)), we need to find the derivative of f(u) with respect to u, and then substitute g(x) for u. Similarly, to find (f∘g)′(x), we need to find the derivative of f(g(x)) with respect to x.
Let's start with finding f ′(u), the derivative of f(u). The derivative of the tangent function, tan(u), is given by:
f ′(u) = sec^2(u)
Now, we substitute g(x) for u:
f ′(g(x)) = sec^2(g(x))
To find (f∘g)′(x), we need to compute the composite function f(g(x)) first. Substituting g(x) = x³, we have:
f(g(x)) = f(x³) = tan(x³)
Next, we find the derivative of tan(x³) with respect to x using the chain rule. The chain rule states that when we differentiate a composite function, we multiply the derivative of the outer function by the derivative of the inner function. Applying the chain rule, we have:
(f∘g)′(x) = sec^2(g(x)) * g'(x)
Substituting g(x) = x³ and g'(x) = 3x²:
(f∘g)′(x) = sec^2(x³) * 3x²
So, the solutions are:
f ′(g(x)) = sec^2(g(x)) = sec^2(x³)
and
(f∘g)′(x) = sec^2(g(x)) * g'(x) = sec^2(x³) * 3x²