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?
Find the second derivative
Find the derivative f'(x)
Find the derivative =
And I'm also stuck at this problem:
Help is greatly appreciated.
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!