Question

Question #12c1e

Answer 1

`3sec^2(3x+2)`

Explanation:

The derivative of the function `tan x` can be found using the quotient rule, and remembering `tan x = (sin x)/(cos x)`:

`f(x) = (g(x))/(h(x)), f'(x) = (g'(x)h(x) - g(x)h'(x))/(h^2(x)`

`tan (x) = sin(x)/cos(x), d/dx tan(x) = ((cos^2x)dx + (sin^2x)dx)/(cos^2(x))`

If we substitute `3x+2` in place of `x`, then because `d/dx (3x+2) = 3`, we multiply by 3 any portion where we took the derivative with respect to x, as indicated by the `dx` placed in the function above.

`d/dx tan(3x+2) = ((3cos^2(3x+2) + 3 sin^2(3x+2))/(cos^2 (3x+2))) = 3(cos^2(3x+2)+sin^2(3x+2))/(cos^2(3x+2) )= (3*1)/cos^2(3x+2) = 3 sec^2(3x+2)`

We perform this last step by recalling our trig identities, namely that `sin^2u + cos^2u = 1`, where `u` is any real, defined expression.

Answer 2

`3 sec^2 (3x + 2)`

Explanation:

`f(x) = tan (3x +2)`

`f’(x) = d/(dx) (tan (3x +2) )`

`u = (3x + 2); du= (3x+2) dx = 3`

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

`f’(x) = sec^2 u .du`

Replacing value of u and du,
`f’(x) = sec^2 (3x+2) * 3 = 3 sec^2(3x + 2)`