`y = (cos x cosec x) / (3+sinx)` `dy/dx ?`

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`y = (cosxcosecx)/(3+sinx)`

This is a derivative of a division.

The derivative of `y = f(x)/g(x)` with respect to x is given by;

`(dy)/dx = (g(x)f'(x)-f(x)g'(x))/(g(x))^2`

`y = (cosxcosecx)/(3+sinx)`

`cosxcosecx = (cosx)xx1/(sinx) = (cosx)/(sinx) = cotx`

`y = (cosxcosecx)/(3+sinx)`

`y = (cotx)/(3+sinx)`

Let;

`f(x) = cotx`

`g(x) = 3+sinx`

`f'(x) = 1/(1+x^2)`

`g'(x) = cosx`

`(dy)/dx `

`= (g(x)f'(x)-f(x)g'(x))/(g(x))^2`

`= (((3+sinx)(1/(1+x^2))-cotxcosx))/(3+sinx)^2`

`= ((3+sinx)(1/(1+x^2))-(cos^2x)/(sinx))/(3+sinx)^2`

`= (3sinx+sin^2x-cos^2x-x^2cos^2x)/((3+sinx)^2sinx(1+x^2))`

We know that `cos2x = cos^2x-sin^2x`

`(dy)/dx `

`= (3sinx-cos2x-x^2cos^2x)/((3+sinx)^2sinx(1+x^2))`

So the answer is `(dy)/(dx)= (3sinx-cos2x-x^2cos^2x)/((3+sinx)^2sinx(1+x^2))`

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