What is the derivative of the following funct ion? (1) y=4sin^-1(3×) (2) y=ln(5×+sin^-1(×) (3) y=3cos(sin^-1(2×+3))

2 Answers
Surya K.
Mar 3, 2018

#(1)# is #4/sqrt(1-x^2)#

#(2)# is #(5sqrt(1-x^2)+1)/((5x+sin^-1(x))sqrt(1-x^2))#

#(3)# is #-(12x+18)/(isqrt(4x^2-12x-10))#

Explanation:

#(1)#:

#d/dx(4sin^-1(x))#

#4*d/dxsin^-1(x)#

#4*1/sqrt(1-x^2)#

#4/sqrt(1-x^2)#

#(2)#:

#d/dx(ln(5x+sin^-1(x)))#

According to the chain rule, #(df)/dx=(df)/(du)*(du)/(dx)#, where #u# is a function within #f#.

Here, we have:

#d/(du)ln(u)*d/dx(5x+sin^-1(x))#

#1/u*(5+1/sqrt(1-x^2))#

But since #u=5x+sin^-1(x)#, we have:

#1/(5x+sin^-1(x))(5+1/sqrt(1-x^2))#

#5/(5x+sin^-1(x))+1/(5x+sin^-1(x)sqrt(1-x^2))#

#(5sqrt(1-x^2)+1)/((5x+sin^-1(x))sqrt(1-x^2))#

#(3)#:

#d/dx(3cos(sin^-1(2x+3)))#

#3*d/dxcos(sin^-1(2x+3))#

Use the chain rule:

#3*d/(du)cos(u)*d/dxsin^-1(2x+3)#

#3*d/(du)cos(u)*d/(dw)sin^-1(w)*d/dx(2x+3)#

#3*(-sin(u))*1/sqrt(1-w^2)*2#

#-6*sin(u)*1/sqrt(1-w^2)#

As #u=sin^-1(2x+3)# and #w=2x+3#, we get:

#-6*sin(sin^-1(2x+3))*1/sqrt(1-(2x+3)^2)#

#-6*(2x+3)*1/sqrt(-4x^2+12x+10)#

#-(12x+18)/(isqrt(4x^2-12x-10))#

dhruv ahlawat
Mar 3, 2018

1) #12/sqrt( 1 - 9x^2)#

2)#(5+ 1/sqrt(1-x^2))/(5x + sin^-1 x)#

3) # -(6 (2x + 3))/ sqrt(1 - (2x + 3)^2)#

Explanation:

You can Solve this Using the Chain Rule
basically,

#d/dx f(g(x))#

#= f'(g(x))* g'(x) #

which is the derivative of the first function(#f(x)#) evaluated at #g(x)#
multiplied by the derivative of #g(x)#

therefore for the first question,
let #f(x) = 4sin^-1(x)#
and #g(x) = 3x#

therefore, for the derivative of #f(g(x))#

derivative of f(x) is #4* 1/sqrt(1-x^2) #
and of g(x) Is 3

therefore , the total derivative is

#4 * 1/sqrt(1 - (3x)^2) * 3#
#= 12 /sqrt(1 - 9x^2) #

2)
similarly, for the second one,
derivative of #ln(x)# is# 1/x# which evaluated at the inside function is #1/ (5x + sin^-1(x))# multiplied by the derivative of the inside function of
#5x + sin^-1(x)#
is # 5 + 1/sqrt(1 - x^2)#

therefore the total derivative is
# (5 + 1/sqrt(1 - x^2)) / (5x + sin^-1(x))#

3)
#dy/dx =d/dx 3 cos(sin^-1(2x+3))#

use the same chain rule twice.
the derivative of the outside function = #-3sin(x)#
which has to be evaluated at the inside function,
#sin^-1(2x+3)#

And the derivative of the inside function found using the chain rule is
#2* 1/ sqrt(1 - (2x + 3)^2)#

therefore,
#-3sin(sin^-1(2x+3)) *2 / sqrt(1 - (2x + 3)^2)#

but since #sin(sin^-1(2x+3)) = 2x+ 3#
the entire derivative is

#-(6(2x+3)) / sqrt(1 - (2x + 3)^2)#

And That's it

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