Question

derivate of secx^2 w.r.t 'x' is??​

Answer 1

Answer:

PLEASE find the attached image.

Step-by-step explanation:

Answer 2

Step-by-step explanation:

$f(x) = {sec \: x}^{2} dx \\ \\ f(x) = \frac{1}{cosx ^{2} } \\ \\ f(x) = cosx ^{ - 2} \\ \\ f'(x) = ( - 2) {cosx}^{ - 3} \frac{d}{dx} {cosx} \\ \\ \: \: \: \: \: \: \: \: \: = (- 2) {cosx}^{ - 3} ( - sinx) \\ \\ \: \: \: \: \: \: \: \: = (- 2 ) \frac{sinx}{ - {cosx}^{3} } \\ \\ \: \: \: \: \: \: = \frac{2sinx}{ {cosx}^{3} } \\ \\ \: \: \: \: \: \: = 2 \: tanx \: {secx}^{2}$