What is the derivative of #arctan(1+2x)^(1/2)#?

Question

2228 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-13T10:43:52

#dy/dx=1/{2(1+x)sqrt(1+2x)}.#

Let #y=arctan(1+2x)^(1/2)#

Put #u=(1+2x)^(1/2)# & #(1+2x)=t.#

With these substitutions, we see that, #y=arctanu, u=t^(1/2), &, t=1+2x...............(1).#

Thus, #y# is a function of #u#, #u# of #t#, &, #t# of #x#.

Accordingly, to find #dy/dx# we need to apply the Chain Rule, which states that in the situation described above,

#dy/dx=dy/(du)*(du)/(dt)*dt/dx#

#={d/(du)(arctanu))}{d/dt(t^(1/2)}{d/dx(1+2x)}#

#={1/(1+u^2)}{(1/2)*t^(-1/2)}(0+2)#

#=1/{1+((1+2x)^(1/2))^2}{1/t^(1/2)}#

#={1/(1+1+2x)}*{1/(1+2x)^(1/2)}#

#=1/(2(1+x))*1/(1+2x)^(1/2)#

#:. dy/dx=1/{2(1+x)sqrt(1+2x)}.#

Do enjoy Maths.!