How do you differentiate #y=cot(1-2x^2)#?

Question

Charles Anctil · Accepted Answer

#y^' = 4x * csc^2(1-2x^2)#

You can differentiate this function by using the chain rule for #cotu#, with #u = 1-2x^2#.

I'll assume that you don't know what the derivative of #cotx# is, but that you do know that

#color(blue)(d/dx(cosx) = -sinx)" "# and #" "color(blue)(d/dx(sinx) = cosx)#

You can derive the derivative of #cotx# by using the quotient rule or by using the product rule and the chain rule. You can find the derivative of #cotx# by using the quotient rule here, so I'll use the product + chain rules.

Start from the fact that

#cot(x) = cosx/sinx#

this can be rewritten as a product of two functions

#cotx = cosx * sin^(-1)x#

If you take #t^(-1)# with #t = sinx#, you can write

#d/dx(cotx) = [d/dx(cosx)] * sin^(-1)x + cosx * [d/dx(t^(-1))]#

#d/dx(cotx) = -color(red)(cancel(color(black)(sinx))) * color(red)(cancel(color(black)(sin^(-1)x))) + cosx * [d/(dt)t^(-1) * d/dx(t)]#

#d/dx(cotx) = -1 + cosx * [-t^(-2) * d/dx(sinx)]#

#d/dx(cotx) = -1 + cosx * (-1/sin^2x * cosx)#

#d/dx(cotx) = -1 -cos^2x/sin^2x#

This can be written as

#d/dx(cotx) = -(overbrace(sin^2x + cos^2x)^(color(blue)(=1)))/sin^2x = -1/sin^2x = -csc^2x#

Your original derivative will thus be

#d/dx(y) = d/(du)(cotu) * d/dx(u)#

#y^' = -csc^2u * d/dx(1-2x^2)#

#y^' = -csc^2(1-2x^2) * (-4x)#

Finally, you get that

#y^' = color(green)(4x * csc^2(1-2x^2))#

Samantha Adkison · Answer

undefined To differentiate y = cot(1 - 2x^2), you use the chain rule.

The derivative of cot(u) with respect to u is -csc^2(u). Thus, the derivative of cot(1 - 2x^2) with respect to (1 - 2x^2) is -csc^2(1 - 2x^2).

Then, by the chain rule, multiply by the derivative of the inner function, which is -4x.

So, the derivative of y = cot(1 - 2x^2) with respect to x is: 
dy/dx = -csc^2(1 - 2x^2) * (-4x) 
     = 4x csc^2(1 - 2x^2).