How do you use the chain rule to differentiate #y=5tan^5(2x+1)#?
1 Answer
# dy/dx = 50(2x+1)^4sec^2(2x+1) #
Explanation:
If you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:
If
# y=f(x) # then# f'(x)=dy/dx=dy/(du)(du)/dx #
I was taught to remember that the differential can be treated like a fraction and that the "
# dy/dx = dy/(dv)(dv)/(du)(du)/dx # etc, or# (dy/dx = dy/color(red)cancel(dv)color(red)cancel(dv)/color(blue)cancel(du)color(blue)cancel(du)/dx) # etc
So with
Using
# dy/dx = (5)(5u^4)(sec^2u)(2) #
# :. dy/dx = 50u^4sec^2u #
# :. dy/dx = 50(2x+1)^4sec^2(2x+1) #