Question

How do you differentiate `sin(arctanx)`?

Answer 1

`1/(1+x^2)^(3/2)`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larr" chain rule"`

`rArrd/dx(sin(arctanx))`

`=cos(arctanx)xxdx(arctanx)`

`=(cos(arctanx))/(1+x^2)`

`=1/(1+x^2)xx1/(sqrt(1+x^2)`

`=1/(1+x^2)^(3/2)`

Answer 2

`1/(x^2+1)^(3/2).`

Explanation:

Let, `y=sin(arc tanx)=sintheta, theta=arc tanx," so that, "tantheta=x.`

Now, `tantheta=x rArr csc^2theta=1+cot^2theta=1+1/tan^2theta,`

`:. csc^2theta=1+1/x^2=(x^2+1)/x^2 rArr sintheta=1/csctheta=x/sqrt(x^2+1).`

Hence, `y=sintheta=x/sqrt(x^2+1).`

Using the Quotient Rule for Diffn.,

`dy/dx={sqrt(x^2+1)d/dx(x)-xd/dx(sqrt(x^2+1))}/{sqrt(x^2+1)}^2...(ast).`

Here, by the Chain Rule,

`d/dx(sqrt(x^2+1))=d/dx(x^2+1)^(1/2),`

`=1/2*(x^2+1)^(1/2-1)*d/dx(x^2+1),`

`=1/2*(x^2+1)^(-1/2)*(2x).`

`rArr d/dx(sqrt(x^2+1))=x/sqrt(x^2+1).`

Utilising this, in `(ast),` we get,

`dy/dx={sqrt(x^2+1)*1-x*x/sqrt(x^2+1)}/(x^2+1),`

`={(x^2+1-x^2)/sqrt(x^2+1)}/(x^2+1),`

`rArr dy/dx=1/(x^2+1)^(3/2),` is the desired Diffn.