F(x) = cos(x) • the integral from 2 to x² + 1 of
e^(u² +5)du
Find F'(x).
When i did this, i got:
-2xsin(x)e^((x²+1)² + 5)
But my teacher got:
-sin(x) • the integral from x² + 1 of e^(u² +5)du + 2xcos(x)e^((x²+1)² + 5)
Do you know why the integral is in his answer? I'm not sure where I went wrong. If you could help, I would greatly appreciate it. Thanks!!
Leibnitz's Rule explains how to take the derivative of an integral. Take a google for it, or consult your textbook.
Basically, you have a product here. cos(x) * Integral f(x)
d/dx of the product is
-sin(x) * Integral + cos(x) * d/dx(Integral(f))
d/dx(Integral) = Integral(df/dx) = f, evaluated at the limits of integration.
To find F'(x), you need to apply the chain rule to the function F(x) = cos(x) times the integral from 2 to x² + 1 of e^(u² + 5)du.
Let's start by finding the derivative of the integral part ∫[2, x²+1] e^(u² + 5)du with respect to x. To do this, we use the fundamental theorem of calculus, which states that if F(x) is the integral of a function f(u), then dF(x)/dx = f(x).
In other words, differentiate the integral's upper limit (x² + 1) with respect to x, and multiply it by the integrand e^(u² + 5):
d/dx [∫[2, x²+1] e^(u² + 5)du]
= e^((x² + 1)² + 5) * d/dx (x² + 1)
Now, let's differentiate the remaining part cos(x) with respect to x:
d/dx [cos(x)] = -sin(x)
Putting it all together, we get:
F'(x) = -sin(x) * e^((x² + 1)² + 5) * d/dx (x² + 1)
Now, let's find d/dx (x² + 1):
d/dx (x² + 1) = 2x
So, substituting this back into our expression, we have:
F'(x) = -sin(x) * e^((x² + 1)² + 5) * (2x)
Simplifying, we get:
F'(x) = -2xsin(x) * e^((x² + 1)² + 5)
Comparing this with your teacher's answer, the difference lies in the presence of the integral term. It seems that your teacher mistakenly forgot to include the integral part when applying the chain rule. The integral should have been differentiated as shown above.
So, the correct answer is indeed:
F'(x) = -2xsin(x) * e^((x² + 1)² + 5)
I hope this clarifies your doubt! Let me know if you have any other questions.