What is the derivative of #(sinx)^(cos^(-1)x)#?

2 Answers
Jul 4, 2017

Answer:

# dy/dx = (sinx)^(cos^(-1)x) \ (cos^(-1)x \ cotx -(ln sinx)/sqrt(1-x^2) )#

Explanation:

Let:

# y=(sinx)^(cos^(-1)x) #

Take Natural logarithms of both sides:

# ln y = ln {(sinx)^(cos^(-1)x) }#
# " " = (cos^(-1)x) \ ln {sinx}#

Differentiate Implicitly,applying the product rule and chain rule:

# 1/y \ dy/dx = (cos^(-1)x)(d/dx ln {sinx}) + (d/dx cos^(-1)x)(ln {sinx}) #
# " " = (cos^(-1)x)(1/sinx \ cosx) + (-1/sqrt(1-x^2))(ln {sinx}) #
# " " = cos^(-1)x \ cosx/sinx -1/sqrt(1-x^2)(ln {sinx}) #
# " " = cos^(-1)x \ cotx -(ln sinx)/sqrt(1-x^2) #

And so:

# dy/dx = y \ (cos^(-1)x \ cotx -(ln sinx)/sqrt(1-x^2) )#
# " " = (sinx)^(cos^(-1)x) \ (cos^(-1)x \ cotx -(ln sinx)/sqrt(1-x^2) )#

Jul 4, 2017

Answer:

# [(sinx)^(cos^(-1)x){sqrt(1-x^2)cotxcos^(-1)x-ln(sinx)}]/sqrt(1-x^2).#

Explanation:

Let, #y=(sinx)^(cos^(-1)x).#

# :. lny=ln{(sinx)^(cos^-1x)}=(cos^(-1)x)(ln(sinx)).#

# :. d/dx(lny)=d/dx{(cos^(-1)x)(ln(sinx))}............(ast).#

Here, by the Chain Rule,

#d/dx(lny)=d/dy(lny)*d/dx(y)=1/y*dy/dx......(ast^1).#

For the Derivative on the R.H.S., we use the Product Rule

and the Chain Rule.

#d/dx{(cos^(-1)x)(ln(sinx))},#

#=cos^(-1)x*d/dx(ln(sinx))+ln(sinx)*d/dxcos^(-1)x,#

#=(cos^(-1)x){1/sinx*d/dx(sinx)}#
#+ln(sinx)*{-1/sqrt(1-x^2)},#

#=(cos^(-1)x)(cosx/sinx)-(ln(sinx))/sqrt(1-x^2),#

#=cotxcos^(-1)x-ln(sinx)/sqrt(1-x^2),#

#={sqrt(1-x^2)cotxcos^(-1)x-ln(sinx)}/sqrt(1-x^2).....(ast^2).#

Using #(ast^1) and (ast^2)" in "(ast),# we have,

#1/y*dy/dx={sqrt(1-x^2)cotxcos^(-1)x-ln(sinx)}/sqrt(1-x^2),#

# rArr dy/dx=[y{sqrt(1-x^2)cotxcos^(-1)x-ln(sinx)}]/sqrt(1-x^2), or,#

#dy/dx=[(sinx)^(cos^(-1)x){sqrt(1-x^2)cotxcos^(-1)x-ln(sinx)}]/sqrt(1-x^2).#

Enjoy Maths.!