How do you find the derivative of # y = cot5x + csc5x #?

Question

Emilia Aguilar · Accepted Answer

#-5csc 5x (cosec 5x-cot 5x)or -5 csc 5x × tan 5x/2# Given #y=cot 5x+csc 5x# #d/dx(cot x)=-csc^2 x# #d/dx(csc x)=-csc x cot x# Now, By chain rule, #d/dx(cot 5x)=-5csc^2 5x# #d/dx(csc 5x)=-5csc 5x cot 5x# Adding, we get, #-5csc 5x (cosec 5x-cot 5x)# which is the answer. Further simplifying, #csc 5x-cot 5x " as " tan (5x/2)# by the identity #csc x-cot x =tan (x/2)# We get the answer as#-5 csc 5x × tan 5x/2#

Adam Adkins · Answer

undefined To find the derivative of $ y = \cot^5(x) + \csc^5(x) $, you can use the chain rule and the derivatives of cotangent and cosecant functions:

$$ \frac{dy}{dx} = \frac{d}{dx}(\cot^5(x)) + \frac{d}{dx}(\csc^5(x)) $$

Applying the chain rule and the power rule for derivatives:

$$ \frac{dy}{dx} = 5\cot^4(x)\frac{d}{dx}(\cot(x)) - 5\csc^4(x)\frac{d}{dx}(\csc(x)) $$

Using the derivatives of cotangent and cosecant:

$$ \frac{dy}{dx} = 5\cot^4(x)(-\csc^2(x)) + 5\csc^4(x)(-\cot(x)\csc(x)) $$

$$ \frac{dy}{dx} = -5\cot^4(x)\csc^2(x) + (-5)\csc^5(x)\cot(x) $$

$$ \frac{dy}{dx} = -5\cot^4(x)\csc^2(x) - 5\csc^5(x)\cot(x) $$

So, the derivative of $ y = \cot^5(x) + \csc^5(x) $ is $ \frac{dy}{dx} = -5\cot^4(x)\csc^2(x) - 5\csc^5(x)\cot(x) $.