How do you use the chain rule to differentiate #y=5tan^5(2x+1)#?

1 Answer
Nov 13, 2016

# 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 "#dx#'s" of a common variable will "cancel" (It is important to realise that #dy/dx# isn't a fraction but an operator that acts on a function, there is no such thing as "#dx#" or "#dy#" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

# 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 # y =5tan^5(2x+1) #, Then:

# { ("Let "u=2x+1, => , (du)/dx=2), ("Let "v=tanu, =>, (dv)/(du)=sec^2u ), ("Let "w=v^5, =>, (dw)/(dv)=5u^4 ), ("Then "y=5w, =>, (dy)/(dw)=5 ) :}#

Using # dy/dx=(dy/(dw))((dw)/(dv))((dv)/(du))((du)/(dx)) # we get:

# dy/dx = (5)(5u^4)(sec^2u)(2) #

# :. dy/dx = 50u^4sec^2u #

# :. dy/dx = 50(2x+1)^4sec^2(2x+1) #