Find f'(c) if f(x)=-3e^(x+2)+ e^(-5)
[NOTE:A small algebraic manipulation is needed first to get f(x) into a form so that the derivative can be taken and your answer isn't a # but rather a formula that returns and answer.]
attempt at the problem:
3e^(x+2)=e^(-5)---->ln both sides
3^(x+2)=-5
f(x)=3^(x+2)+5
so then: f'(x)=3^(x+2)*ln(3)
This answer is incorrect and I don't know what else to do. I also tried to just take the derivative without manipulating the problem and my answer was still incorrect. Please help. Thank you!!
-3e^(x+2)=e^(-5)
e-3(x+2)=e-5
now take the ln of both sides.
after you ln both sides you end up with
-3(x+2)=-5
-3x-6=-5
f(x)= -3x-1
f'(x)= -3
unfortunately that was not correct either, but I appreciate your help! :)
To find the derivative, let's start with the original function:
f(x) = -3e^(x+2) + e^(-5)
To differentiate this function, we'll use the rules of differentiation. The first term, -3e^(x+2), is the product of a constant (-3) and a function (e^(x+2)). According to the product rule, the derivative of the product of two functions is given by:
(d/dx)(c * f(x)) = c * (d/dx)(f(x)) + f(x) * (d/dx)(c)
In this case, c = -3 and f(x) = e^(x+2). We need to find the derivative of e^(x+2) and multiply it by -3, as well as find the derivative of -3 and multiply it by e^(x+2).
Let's start by finding the derivative of e^(x+2). The derivative of e^x with respect to x is simply e^x, assuming the base of the exponential is e. However, in this case, we have (x+2) inside the exponent, which makes it more complicated.
To differentiate e^(x+2), we'll apply the chain rule. The chain rule states that if we have a composite function, u(v(x)), then the derivative is given by:
(d/dx)(u(v(x))) = (du/dv)(v(x)) * (dv/dx)
In this case, u(v(x)) = e^v and v(x) = x + 2. Therefore, we have:
(d/dx)(e^(x+2)) = (d/du)(e^u) * (d/dx)(x+2)
The derivative of e^u with respect to u is just e^u. The derivative of x+2 with respect to x is simply 1. Therefore,
(d/dx)(e^(x+2)) = e^(x+2) * 1 = e^(x+2)
Now, let's find the derivative of -3. The derivative of a constant is always zero, so:
(d/dx)(-3) = 0
Now, we can put everything together to find the derivative of f(x):
f'(x) = -3 * e^(x+2) + e^(-5) * e^(x+2)
To simplify this expression, we need to combine the two terms with the same base, e. Using the property a^c * a^d = a^(c+d), we can combine the two exponential terms:
f'(x) = (-3 + e^(-5)) * e^(x+2)
However, we want to find f'(c), not f'(x). To find the derivative at a specific point c, we substitute c into the expression for f'(x):
f'(c) = (-3 + e^(-5)) * e^(c+2)
So, the final answer for f'(c) is:
f'(c) = (-3 + e^(-5)) * e^(c+2)