Your reasoning is correct. Try the substitution to make it even clearer.
I have to take the AP Calculus AB test tomorrow, so anyone who can clarify this for me quickly would be much, much appreciated.
I understand that when I perform the chain rule with a derivative, such as d/dx (2x+1)^3, I must take the derivative, yielding 3(2x+1)^2, and then I must multiply by the derivative of the inside, d/dx (2x+1) = 2, yielding 2*3(2x+1)^2 -> 6(2x+1)^2.
However, I am confused on this when I am using something other than a simple binomial. If I must evaluate d/dx (e^2x), would the derivative simply be (e^2x), or would I multiply that by the derivative of the exponent (d/dx 2x = 2), making it 2*(e^2x) -> (2e^2x)?
If the latter is true and I must multiply by the derivative of the exponent, then what do I do when I must take the antiderivative of it? For example, if I am given ∫(e^2x)dx, would the answer be (1/2)(e^2x)+C ?
Please correct anything I am mistaken on. Thank you.
Yes, any time you have a "differentiation rule" for f(x) but have some function of x, say u(x), you must use the chain rule:
In this particular example, u(x)= 2x so
A slightly more complicated example is:
Here so
That's exactly what "substitution" in integration is. To integrate \int e^{2x}dx[/tex], let u= 2x. then du= 2dx so (1/2)du= dx. Replacing 2x by u and dx by (1/2)du, the integral becomesIf the latter is true and I must multiply by the derivative of the exponent, then what do I do when I must take the antiderivative of it? For example, if I am given ∫(e^2x)dx, would the answer be (1/2)(e^2x)+C ?
[tex]\int e^{2x}dx= \int e^u ((1/2)du)= (1/2)\int e^u du= (1/2)e^u+ C= (1/2)e^{2x}+ C[//math]
just as you say.
HOWEVER, integrating is harder than differentiating and this is one reason. If the integral were
we might try the substitution . Then du= (1/2)x dx. If we do the same as before, 2du/x= dx and when we replace by u and dx by 2du/x we get
and still have x in the integral. We can't take it out of the integral as we did with the "2" in the first integral because it is a variable and not a constant. That is why
important in the "Normal Probability Distribution", cannot be written in terms of elementary functions.
Notice that, even though it seems more difficult, the integral
CAN be done easily. Using the substitution now, , [tex] (1/2)du= x dx and the integral becomes
Please correct anything I am mistaken on. Thank you.