Find the derivative dy/dx of:
y=(2x^3-4x+1)^-6
The answer is:
-12(3x^2-2)(2x^3-4x+1)^-7
I'm able to go as far as:
y=(2x^3-4x+1)^-6
u=(2x^3-4x+1)
y=u^-6
dy/du= -6u^-7
du/dx= (6x^2-4)
dy/dx=(dy/du)(du/dx)
dy/dx (y)= -6(2x^3-4x+1)^-7(6x^2-4)
I can't get beyond that, however. If you can enlighten me as to what I'm doing wrong, I'll let you be co-ruler when I take over the world.

Question

Find the derivative dy/dx of:
y=(2x^3-4x+1)^-6
The answer is:
-12(3x^2-2)(2x^3-4x+1)^-7
I'm able to go as far as:
y=(2x^3-4x+1)^-6
u=(2x^3-4x+1)
y=u^-6
dy/du= -6u^-7
du/dx= (6x^2-4)
dy/dx=(dy/du)(du/dx)
dy/dx (y)= -6(2x^3-4x+1)^-7(6x^2-4)
I can't get beyond that, however.  If you can enlighten me as to what I'm doing wrong, I'll let you be co-ruler when I take over the world.

anonymous · Answer

what makes you think you are doing something wrong? Looks fine.

anonymous · Answer

$$\frac{dy}{dx} = -6\frac{(6x^2-4)}{(2x^3-4x+1)^7} = -6\times2 \frac{(3x^2-2)}{(2x^3-4x+1)^7}=-12\frac{(3x^2-2)}{(2x^3-4x+1)^7}$$