Answer:
[tex]2sec^2xtanx[/tex]
Step-by-step explanation:
[tex]\frac{d}{dx} tan^2x[/tex]
[tex]\frac{d}{dx} tanxtanx[/tex]
[tex](\frac{d}{dx}tanx)(tanx)+(tanx)(\frac{d}{dx}tanx)[/tex]
[tex]\frac{d}{dx}tanx=\frac{d}{dx}\frac{sinx}{cosx}=\frac{(cosx)(cosx)-(sinx)(-sinx)}{(cosx)^2}=\frac{cos^2x+sin^2x}{cos^2x}=\frac{1}{cos^2x}=sec^2x[/tex]
[tex](sec^2x)(tanx)+(tanx)(sec^2x)[/tex]
[tex]2sec^2xtanx[/tex]
Helpful tips:
Product Rule: [tex]\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)=\frac{d}{dx}f(x)*g(x)+f(x)*\frac{d}{dx}g(x)[/tex]
Quotient Rule: [tex]\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}=\frac{g(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}g(x)}{(g(x))^2}[/tex]
Pythagorean Identity: [tex]cos^2x+sin^2x=1[/tex]
