Question

What is the derivative of ( tanx+cotx)/(tanx-cotx)Answer is: `-2sec2x tan2x`. How to get this answer??Please help me

What is the derivative of ( tanx+cotx)/(tanx-cotx)Answer is: -2sec2x tan2How to get this answer.Please help me

Answer 1

`=-2sec 2x tan 2x`

Explanation:

Let's Apply the Quotient Rule here.

[The Quotient Rule States That if `f(x) = g(x)/(h(x)), then`

`d/dxf(x) = (g'(x)h(x) - g(x)h'(x))/(h(x))^2`]

Let's assume `f(x) = (tan x + cotx )/(tanx - cotx)`

So, `d/dxf(x) = d/dx ((tanx + cotx)/(tanx - cotx))`

`= ((d/dx(tanx + cotx))xx (tanx - cotx) - (tan x + cot x) xx d/dx(tan x - cotx))/((tanx - cot x)^2)`

`= (((sec^2 x - csc^2x) xx (tanx - cot x)) - ((tanx + cotx)xx(sec^2x + csc^2x)))/((tanx -cotx)^2)`

`= (cancel(sec^2xtanx) -csc^2xtanx - sec^2xcotx cancel(+ csc^2xcotx) cancel(- sec^2xtanx) - sec^2xcotx - csc^2xtanx cancel(- csc^2xcotx))/((tanx -cotx)^2)`

`= (-2(csc^2xtanx + sec^2xcotx))/((tanx -cotx)^2)`

`= (-2(1/(sinx cosx) + 1/(sin x cosx)))/((tanx -cotx)^2)`

`= (-4/(sinxcosx))/((sin^2x - cos^2x)/(sinx cosx))^2`

`= ((-4/(sinxcosx)) xx (sin^2xcos^2x))/(cos^2(2x))`

`= (-2sin 2x)/(cos^2 2x)`

`= (-2 xx sin (2x)/cos (2x) xx 1/cos (2x))`

`= -2tan 2x sec 2x`

Finally Got The Answer Right. Thanks Everyone Who helped me!

And, Hope this Helps.

Answer 2

`d/dx (tanx+cotx)/(tanx-cotx) = -2sec2xtan2x`

Explanation:

For convenience, define:

`y = (tanx+cotx)/(tanx-cotx)`

Before we differentiating we can simplify the expression:L

`y = (sinx/cosx+cosx/sinx)/(sinx/cosx-cosx/sinx)`

`\ \ = ((sin^2x+cos^2x)/(sinxcosx))/((sin^2x-cos^2x)/(sinxcosx))`

`\ \ = ((sin^2x+cos^2x)/(sinxcosx))/((sin^2x-cos^2x)/(sinxcosx))`

`\ \ = (sin^2x+cos^2x)/(sinxcosx) * (sinxcosx)/(sin^2x-cos^2x)`

`\ \ = (sin^2x+cos^2x)/(sin^2x-cos^2x)`

And we can use two fundamental trigonometric identities:

`sin^2x+cos^2x -= 1 \ \ \ \ \ \ \ \ \ \` (Pythagorean Identity)
`cos2x -= cos^2x - sin^2x \ \ ` (Multiple Angle Identity)

Giving us

`y = (1)/(-cos2x) = -sec2x`

And now we differentiate wrt `x` using the chain rule to get:

`dy/dx = (-sec2xtan2x)(2)`

Thus:

`d/dx (tanx+cotx)/(tanx-cotx) = -2sec2xtan2x \ \ \ ` QED