Question

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

Answer 1

`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)`