Question

Use the chain rule to find the derivative of the following function.
f(x) = ln(2x^2 - 6)

Answer 1

Remember that d(ln(u)/dx = (1/u)*du/dx. That implies that we have:
f'(x) = (1/(2x^2 - 6))*(4x) = 4x/(2x^2 - 6) = 2x/(x^2 - 3)

Answer 2

yes I agree with bmp :)

Answer 3

\[f(x) = \ln (2x^2 - 6)\]\[u = 2x^2 - 6\]\[f(x) = \ln u\]\[f'(x) = \frac{dy}{du}\frac{du}{dx}\]\[f'(x) = \frac{d}{du}\left( \ln u \right)\frac{d}{dx}\left( 2x^2 - 6 \right)\]\[f'(x) = \frac{1}{u}\left( 4x \right)\]\[f'(x) = \frac{4x}{2x^2 - 6}=\frac{2x}{x^2 - 3}\]

Answer 4

thank you both very much

Answer 5

Or, raise both LHS and RHS to the power of 'e' and differentiate, taht 'll be way easier I guess.