How do you differentiate #sin(arctanx)#?

2 Answers
Jim G.
Aug 28, 2017

#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)#

Ratnaker Mehta
Aug 29, 2017

# 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.

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