Find the derivative of the function h(t) = (t^3/(t^7+3))^2

Question

The problem is: $h(t) = (\frac{t^3}{(t^7+3)})^2$

I got as an answer enter image description here or 2\frac{t^3}{t^7+3}\left(\frac{\left(t^7+3\right)\left(3t^2\right)-7t^6}{\left(t^7+3\right)^2}\right) It was incorrect though.

Thanks for any help!

Answer 1

Apply the power so you have $$\frac {t^6}{(t^7+3)^2}$$ Then apply the quotient rule $$\frac{6t^5(t^7+3)^2-2(t^7+3)(7x^6)x^6}{(t^7+3)^4}$$ $$\frac{6t^5(t^7+3)^2-14t^{12}(t^7+3)}{(t^7+3)^4}$$