If y = e^cosec x, how do I differentiate (dy/dx)?

take ln of both sides

ln y = cosec x

d/dx(ln y) = d/dy(ln y) dy/dx = 1/y dy/dx

d/dx(cosec x) = - cot x cosec x

so 1/y dy/dx = - cot x cosec x

or dy/dx = -y cot x cosec x

= - e^cosec x cosec x cot x

Heres how you answer it go through it clearly

ln y = cosec x

d/dx(ln y) = d/dy(ln y) dy/dx = 1/y dy/dx

d/dx(cosec x) = - cot x cosec x

then you use this final step dy/dx = -y cot x cosec x

= - e^cosec x cosec x cot x

to get dy/dx of this equation, you don't need to get the ln of both sides since logarithmic differentiation is used only when the base and it's exponent are both not constant, and e is a constant with a value of 2.718282...

so here's how to manipulate it:

the derivative of e raised to a certain function (represented here as u) is e^u multiplied by the derivative of u.

so now, the equation can be written as

dy/dx = e^cscx (-cscxcotx), since the derivative of cscx is -cscxcotx.

simpifying it, the final answer would be:

dy/dx = -e^cscx (cotxcscx)

i'm fresh from differential calculus so i'm pretty sure of this answer. :)

take the ln of both sides..

ln y= ln (e^cosec x)...

making it..

ln y= cosec x (ln of e=1)

then get dy/dx

D (ln y)= D (cosec x)

(1/y) dy/dx= - (cosec x)(cot x)

dy/dx= -y(cosec x)(cot x)

y = e^(cosec(x))

Let X = cosec(x)

then y = e^X

y' = X'e^X

By the product rule: X' = -cos(x)/sin²(x) = -cosec(x)cot(x)

so y' = -cosec(x)cot(x)e^(cosec(x))