1.) can be written as f(x) = x^(-2) - x^(1/3),
so the derivative will be:
f '(x) = -2x^(-3) - (1/3)x^(-2/3)
2.) you can you the quotient rule, where dy/dx = [V(du/dx) - U(dv/dx)] / V^2 and u = x^2, v = (1-3x)
Hey guys. I'm having trouble with derivatives and I need someone's help with my homework. Could you guys please explain to me how to solve these problems?
1.
Given
Find the second derivative
2.
Given
Find the derivative f'(x)
3.
Given
Find the derivative =
And I'm also stuck at this problem:
4.
Given
Find
Help is greatly appreciated.
Thanks guys!
--RS
Sure. First of all, the derivative of a sum of functions is the sum of derivatives of those functions, so:
y = f(x) + g(x) + h(x)
dy/dx = f '(x) + g '(x) + h '(x)
The power rule tells us that for a function like x^n:
dy/dx = nx^(n-1)
In other words we bring the power down in front of x, and then decrease the power by one.
when you have the inverse of a function x^n you have 1/ x^n, and this can also be written as x^-n. For the first term I did it like this, and then used the power rule, to get -nx^(-n-1).
For the second term you can wite square roots as fractions. For example sqrt(x) can be written x^(1/2), the cubed root can be written x^(1/3), and the nth root can be written x^(1/n). Once you have written it like this you can simply apply the power rule to it again. so than if f(x)= nth root of x:
f(x) = x^(1/n)
dy/dx = (1/n)x^(1/n - 1)
When you have a problem like this, it's much easier to write inverses out as negative powers, and square roots out as fractional powers first. Knowing this, you could probably make a good attempt at the second derivative? If not I'll try it for you. Cheers!