Question

e^cosec^2x find the derivative of this equation

Answer 1

From the definition of cosec, we gave: cosec(2x)=1/sin(2x)

Use the chain rule, namely:

(1/u(x))’=-u’/u^2

Here: u=sin(2x), hence

u’=2cos(2x)

therefore

(1/sin(2x))’=-2cos(2x)/(sin(2x))^2

Using the relation: cos(2x)/sin(2x)=cot(2x) leads to

(cosec(2x))’=-2cot(2x).cosec(2x)

Answers may vary, obviously

Answer 2

Let y = $e^x \csc^2x$

We have to find, the derivative of $e^x \csc^2x$ is:

Solution:

y = $e^x \csc^2x$

Using the identity:

$\dfrac{d(uv)}{dx} =u\dfrac{dv}{dx} +v\dfrac{du}{dx}$

$\dfrac{dy}{dx} =e^x\dfrac{d(\csc^2x)}{dx} +\csc^2x\dfrac{d(e^x)}{dx}$

⇒ $\dfrac{dy}{dx} =e^x(-2\cot x\csc^2 x)+e^x\csc^2x$

⇒  $\dfrac{dy}{dx} =e^x\csc^2x(1-2\cot x)$

∴ $\dfrac{dy}{dx} =e^x\csc^2x(1-2\cot x)$

Thus, the derivative of given equation is "$e^x\csc^2x(1-2\cot x)$".