## Calculus (3rd Edition)

y'=2(x+2)$sec^{2}$($x^{2}$+4x)
The chain rule states that if y=g(h(x)), then y'=g'(h(x))h'(x). Since y=tan($x^{2}$+4x), we can set g(x)=tan(x) (outside function) and h(x)=$x^{2}$+4x (inside function), and use the chain rule to find y'. Therefore, y'=g'($x^{2}$+4x)h'(x) We know that $\frac{d}{dx}$[tan(x)]=$sec^{2}$(x), which can be found by using the quotient rule to find the derivative of the function g(x)=$\frac{sin(x)}{cos(x)}$, which of course is equal to tan(x) $\frac{d}{dx}$[$x^{2}$+4x]=2x+4, using the power rule Now that we know that g'(x)=$sec^{2}$(x) and h'(x)=2x+4, we can find y'. y'=$sec^{2}$($x^{2}$+4x)*(2x+4) =2(x+2)$sec^{2}$($x^{2}$+4x)