Question

FIND dy/dx ✌️



PLZ GUYS​

Answer 1

♦To Find♦:

The derivative of y given as:

$y = log_{ \sin(x) }( \sec(x) ) + 10x {}^{2}$

.

.

Solution:

$\frac{d}{dx} [ log_{ \sin(x) }( \sec(x) ) + 10x {}^{2} ]$

can be written as:

$y' = \frac{d}{dx} [ log_{ \sin(x) }( \sec(x) ) ] + \frac{d}{dx} [10 {x}^{2} ]$

Here,

(1)

$\frac{d}{dx}[ log_{ \sin(x) }( \sec(x) ) ] = \frac{ln( \sec(x)) }{ln( \sin(x)) }$

Now, on using the quotient rule, we get:

$\frac{d}{dx}[f {}^{1} ] = \frac{ \frac{d}{dx} [ln( \sec(x) )].ln( \sin(x) ) - ln( \sec(x) ). \frac{d}{dx} [ln( \sin(x)) ]</p><p>}{(ln( \sin(x) ) {}^{2} }$

$\frac{d}{dx} [f {}^{1} ] = \frac{ \sin {}^{2} ( x ).ln( \sin(x) ) - \cos {}^{2} (x) .ln( \sec(x) ) }{ \cos( x ). \sin(x) .ln( \sin {}^{2} (x) )}$

And,

(2)

$\frac{d}{dx}[f {}^{2} ] = \frac{d}{dx} [10 {x}^{2} ] = 20x$

Therefore,

$\frac{dy}{dx} = \frac{ \sin {}^{2} ( x ).ln( \sin(x) ) - \cos {}^{2} (x) .ln( \sec(x) ) }{ \cos( x ). \sin(x) .ln( \sin {}^{2} (x) )} + 20 {x}^{2}$