Question

If y=2sinx+4secx-4cotx, then find $\frac{dy}{dx}$

• A. $2cosx+4ta{n}^{2}x-4cosecx$
• B. $-2cosx+4sextanx-4cose{c}^{2}x$
• C. $2cosx+4secxtanx+4cose{c}^{2}x$
• D. none of these

Answer 1

The correct option is C $2cosx+4secxtanx+4cose{c}^{2}x$

y= 2sinx + 4secx - 4cotx
$\frac{dy}{dx}=\frac{d(2sinx+4secx-4cotx)}{dx}$
=$\frac{d\left(2sinx\right)}{dx}+\frac{d\left(4secx\right)}{dx}+\frac{d(-4cotx)}{dx}$
=$\frac{2dsinx}{dx}+\frac{4d\left(secx\right)}{dx}-\frac{4d\left(cotx\right)}{dx}$
$=2cosx+4secxtanx-4(-cose{c}^{2}x)$
=$(since\frac{dsinx}{dx}=cosx;\frac{dsecx}{dx}=secxtanx;\frac{dcotx}{dx}=-cose{c}^{2}x)$
$\frac{dy}{dx}=2cosx+4secxtanx+4cose{c}^{2}x$