Question

Question #a4762

Answer 1

`d/dx(tanx)=sec^2x`

Explanation:

Using the `color(blue)"trigonometric identities"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(tanx=(sinx)/(cosx))color(white)(2/2)|)))`

differentiate using the `color(blue)"quotient rule"`

`"Given " f(x)=(g(x))/(h(x))" then"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(2/2)|)))`

`"here "g(x)=sinxrArrg'(x)=cosx`

`"and "h(x)=cosxrArrh'(x)=-sinx`

`rArrf'(x)=(cosx(cosx)-sinx(-sinx))/(cos^2x)`

`color(white)(rArrf'(x))=(cos^2x+sin^2x)/(cos^2x)`

`• cos^2x+sin^2x=1" and " 1/(cos^2x)=sec^2x`

`rArrd/dx(tanx)=1/(cos^2x)=sec^2x`

This result is a `color(blue)"standard derivative"` and worth remembering.

`color(blue)"Note:"` This is a Calculus question not Algebra