Calculus question

Question

1. Let 𝑔(𝑥)= 𝑥^3 + 𝑒^𝑓(𝑥), 𝑓(0)=7, and 𝑓′(0)=−9. Find 𝑔′(0).

g'(0)=

1. Now let ℎ(𝑢)=𝑠𝑖𝑛(𝑢)

𝑑𝑑𝑥∣∣𝑥=0(ℎ∘𝑔)(𝑥)=

Answer 1

Greetings to you, Jason! Great questions! Let me help along the way!

1) Before I start solving these problems, let me get these facts straight:

The derivative of x ^ (n) is n * x ^ (n - 1).

The derivative of e ^ x is e ^ x.

Let's take the derivatives term by term. The derivative of x ^ 3 is 3 * (x ^ 2).

The derivative of e ^ ( f (x) ) is NOT e ^ ( f(x) ) because the exponent is not x. We must use the Chain Rule to multiply the effect of another function (of f(x)).

The Chain Rule tells us that the derivative of the function j( g(x) ) IS j ' ( k(x) ) * k ' (x). In this case, let j (x) = e ^ x and k(x) = f(x)

The derivative of e ^ ( f(x) ) is e ^ ( f(x) ) * f ' (x).

Combining everything makes g ' (x) = 3 * (x ^ 2) + ( e ^ f(x) ) * f ' (x).

Substituting x = 0 makes it g ' (0) = 3 * (0 ^ 2) + ( e ^ f(0) ) * f ' (0) = ( e ^ 7 ) * ( -9 ) <<< (ANSWER)

2) If we want to know what the derivative of h ( g(x) ) will be, use the Chain Rule (look at the previous problem). In this context, d/dx ( h( g(x) ) ) = h ' ( g(x) ) * g ' (x).

Therefore, at x = 0, d/dx ( h( g(0) ) ) = h ' ( g(0) ) * g ' (0)

g(0) = 0 ^ 3 + e ^ ( f(0) ) = e ^ 7 and g ' (0) = (-9) * (e ^ 7)

If h (x) = sinx, then h ' (x) = cosx (by the rules of trignometry)

Therefore, combining all of this information leads to:

d/dx ( h( g(0) ) ) = h ' ( g(0) ) * g ' (0) = cos ( e ^ 7 ) * -9 * (e ^ 7) <<<<<<<<< (ANSWER)

Hope this helped you! Please let me know if you have additional questions!