Question

Y is equal to tan inverse cos 5x + sin 5x upon cos 5x - sin 5x find derivative ​

Answer 1

Answer:

take cos5x common , use Tan(A+B)

Answer 2

The required derivative of $y=tan^{-1}(\frac{cos5x+sin5x}{cos5x-sin5x} )$ is $\frac{dy}{dx}=5$ .

Given :

The expression $y=tan^{-1}(\frac{cos5x+sin5x}{cos5x-sin5x} )$

To Find :

The derivation of y.

Solution :

We can find the solution to this problem in the following way.

We can rewrite the given expression.

$y=tan^{-1}[\frac{cos5x+sin5x}{cos5x-sin5x} ]\\y=tan^{-1}[\frac{cos5x(1+\frac{sin5x}{cos5x} )}{cos5x(1-\frac{sin5x}{cos5x})} ]$

We can now cancel the common term in the numerator and the denominator and also use the following trigonometric formula.

$tanA=\frac{sinA}{cosA}$

We get the following resultant equation.

$y=tan^{-1}[\frac{cos5x(1+\frac{sin5x}{cos5x} )}{cos5x(1-\frac{sin5x}{cos5x})} ]\\y=tan^{-1}[\frac{1+tan5x}{1-tan5x} ]\\y=tan^{-1}[\frac{tan(\frac{\pi}{4}) +tan5x}{1-tan(\frac{\pi}{4})tan5x} ]$

We now use the following trigonometric equation.

$tan(A+B)=\frac{tanA +tanB}{1-tanAtanB}$

We get the following simplified form.

$y=tan^{-1}[\frac{tan(\frac{\pi}{4}) +tan5x}{1-tan(\frac{\pi}{4})tan5x} ]\\y=tan^{-1}[tan(\frac{\pi}{4}+5x)]\\y=\frac{\pi}{4}+5x$

We now differentiate y in the following way.

$\frac{dy}{dx}=\frac{d}{dx}(\frac{\pi}{4}+5x)\\\frac{dy}{dx}=\frac{d}{dx}(\frac{\pi}{4})+\frac{d}{dx}(5x)\\\frac{dy}{dx}=0+5\\\frac{dy}{dx}=5$

The derivative of $y=tan^{-1}(\frac{cos5x+sin5x}{cos5x-sin5x} )$ is $\frac{dy}{dx}=5$ , which is the required answer.

#SPJ2