Hi, I have a calculus test tomorrow and I understand how to complete all of the problems, but I am unsure behind the reasoning of taking two derivatives. Can someone please explain the following:

1. g(x)=e^(f(x))
g'(x)=e^(f(x))f'(x)

2. f(x)=Integral from 0 to 3x of the square root of (4+t^2)dt
f'(x) = 3 times the square root of (4+9x^2).

For the second derivative, I know it has something to do with the 2nd fundamental theorem of calculus, but if you could elaborate I would appreciate it greatly.

Thank you very much.

1) d/dx e^(u) is always (u')(e^(u)). In other words, take the original function and multiply it by the derivative of u (f(x)).

2) The 2nd FTC: Take the b (top number of the integral) and plug that in for t. Multiply by the derivative of b. b'(f(b)).

Sure! Let me explain the reasoning behind taking two derivatives in each of these cases.

1. To find the derivative of g(x) = e^(f(x)), we use the chain rule. The chain rule states that if we have a composition of functions, like in this case where we have e^u and u = f(x), the derivative is given by the product of two derivatives: the derivative of the outer function (e^u) with respect to u, multiplied by the derivative of the inner function (f(x)) with respect to x.

So, applying the chain rule, we take the derivative of g(x) = e^(f(x)) as follows:

g'(x) = (e^u) * (du/dx),

where u = f(x). The derivative of e^u with respect to u is simply e^u. And the derivative of f(x) with respect to x is f'(x). Therefore, the derivative of g(x) is:

g'(x) = e^(f(x)) * f'(x).

This gives us the relationship between the derivative of g(x) and the derivatives of e^(f(x)) and f(x).

2. To find the derivative of f(x) = Integral from 0 to 3x of the square root of (4+t^2)dt, we use the second fundamental theorem of calculus. This theorem relates differentiation and integration, and states that if we have an integral with a variable upper limit, like in this case where the upper limit is 3x, then the derivative of the integral with respect to x is equal to the integrand evaluated at the upper limit (in this case 3x), multiplied by the derivative of the upper limit with respect to x.

So, to find f'(x), we apply the second fundamental theorem of calculus:

f'(x) = (square root of (4+(3x)^2)) * (d(3x)/dx).

The derivative of 3x with respect to x is simply 3. Simplifying further, we get:

f'(x) = (square root of (4+9x^2)) * 3.

This gives us the relationship between the derivative of f(x) and the integrand square root of (4+t^2), with the upper limit 3x evaluated at 3x, multiplied by the derivative of the upper limit with respect to x.

I hope this helps clarify the reasoning behind taking two derivatives in these cases. Good luck with your calculus test!