Question
Can anyone explain how to use the chain rule and power rule together to find the first derivative and please do these examples.
f(x) = -3x^4/(sqrt(4x-8))
g(x) = ((2x+5)/(6-x^2)^4
g(x) = (8x^3)(4x^2+2x-3)^5
y = [1/(4x+x^2)^3]^3
Thank you so much!
f(x) = -3x^4/(sqrt(4x-8))
g(x) = ((2x+5)/(6-x^2)^4
g(x) = (8x^3)(4x^2+2x-3)^5
y = [1/(4x+x^2)^3]^3
Thank you so much!
Answers
I will give you the 'first-line' derivatives of the first two questions
a) f'(x) = [(4x-8)^(1/2)(-12x^3) - (-3x^4)(1/2(4x-8)^(-1/2)(4)]/(4x-8)
= ....
b) I see you had a double bracket at the front, but one bracket is not closed.
Did you mean [(2x+5)/(6-x^2)]^4 ?
c) g(x) = (8x^3)(4x^2+2x-3)^5
g'(x) = 5(4x^2+2x-3)^5(8x+2)(8x^3) + 24x^2(4x^2+2x-3)^5
= ....
d) y = [1/(4x+x^2)^3]^3
y = (4x+x^2)^-9
y' = -9(4x + x^2)^-10(4+2x)
= ....
a) f'(x) = [(4x-8)^(1/2)(-12x^3) - (-3x^4)(1/2(4x-8)^(-1/2)(4)]/(4x-8)
= ....
b) I see you had a double bracket at the front, but one bracket is not closed.
Did you mean [(2x+5)/(6-x^2)]^4 ?
c) g(x) = (8x^3)(4x^2+2x-3)^5
g'(x) = 5(4x^2+2x-3)^5(8x+2)(8x^3) + 24x^2(4x^2+2x-3)^5
= ....
d) y = [1/(4x+x^2)^3]^3
y = (4x+x^2)^-9
y' = -9(4x + x^2)^-10(4+2x)
= ....