Let g(x)=(3x+x^2)(2+e^(x−1)). Find g′(1)
So I found the derivative
(2x+3)(e^(x−1)+2)+(x^2+3x)e^(x−1)
I tried plugging in 1 but I'm getting a different answer..
Is 19 the right answer?
(2*1+3)(1+2)+(1^2+3*1)1 = 5*3 + 1+3 = 19
To find the derivative of g(x)=(3x+x^2)(2+e^(x−1)), you correctly applied the product rule. However, it seems that there might be an error while evaluating the derivative at x=1. Let's go through the steps again to find g'(1).
First, let's rewrite the derivative expression:
g'(x) = (2x+3)(e^(x−1)+2) + (x^2+3x)e^(x−1)
Now, plug in x=1 into the derivative expression:
g'(1) = (2(1)+3)(e^(1−1)+2) + ((1)^2+3(1))e^(1−1)
Simplifying this expression, we have:
g'(1) = (2+3)(e^0+2) + (1+3)(e^0)
g'(1) = (5)(1+2) + (4)(1)
g'(1) = 5(3) + 4(1)
g'(1) = 15 + 4
g'(1) = 19
Therefore, g'(1) = 19.