hello
I need some help with finding the derivative of 4x(squareroot 1-x^2)
Please look at the drawing, the correct answer is in green ,I cant seem to get it?
Please help
Thanks
Hi wolfhound,
just use a combination of the product rule and chain rule to solve.
$\displaystyle \frac{d}{dx}4x\sqrt{1-x^2}=4x\frac{d}{dx}\sqrt{1-x^2}+\sqrt{1-x^2}\frac{d}{dx}4x$
$\displaystyle =(4x)0.5(1-x^2)^{-0.5}(-2x)+\sqrt{1-x^2}(4)$
$\displaystyle =\frac{-4x^2}{\sqrt{1-x^2}}+\frac{4(1-x^2)}{\sqrt{1-x^2}}$
since $\displaystyle \frac{\sqrt{1-x^2}\sqrt{1-x^2}}{\sqrt{1-x^2}}=\frac{1-x^2}{\sqrt{1-x^2}}$