Calculus – Differentiation and Integration

[Jeffrey Chudy]

Calculus – Differentiation and Integration

[Anonymous]

Hi,

I was working through the handbook and had a couple questions about the examples given.

if you have y= 1/(x+1) and are asked “calculate dy/dx at point x=2”, I understand this is equal to f'(2).

Question 1:
To answer the question do you:

a) substitute 2 into the equation, y=1/(2+1) or y=1/3. Then dy/dx = 0
b) since y=1/(x+1) => y=(x+1)^-1, therefore y’= (-x-1)^-2

Question 2: (Integration)
The handbook gives the example:
The integral of e^3x I’m not sure how to get to the result y’=3e^3x (chain rule).

Also, if I have written any of the notation wrong, I would appreciate suggestions on how to fix it.

Thanks!

[Anonymous]

You need to first differentiate f(x):

f'(x) = – 1 / (x + 1)^2
f'(2) = – 1/9

Integration is sometimes guesswork. In the case of the integral of e^3x, you need to find a function F(x) which when differentiated produces e^3x. So if I guess F(x) = e^3x, then F'(x) = 3e^3x. So it appears than F'(x) is 3 times more than e^3x. So I will guess F(x) = e^3x / 3, then F'(x) = 3e^3x / 3 = e^3x. So the integral of e^3x is e^3x / 3.

I hope this helps.

[Author]

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