How do you find the derivative of #sin^-1(2x+1)#?

2 Answers
Nov 2, 2016

The answer is #2/sqrt(1-(2x+1)^2)#

Explanation:

For this equation you would use the [chain rule] (https://socratic.org/calculus/basic-differentiation-rules/chain-rule) so you take the derivative of the outside:
#(sin^-1)#
times the derivative of the inside:
#(2x + 1)#

So the derivative of #sin^-1# otherwise known as #arcsin# is #1/sqrt(1-x^2)#
Differentiating Inverse Trigonometric Functions

but in this case #(2x-1)# is acting as #x# so it's
#1/sqrt(1-(2x-1)^2)#

Next the derivative of #2x-1# is #2#

So the answer becomes outside times inside
Which is

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

Here are the derivatives of inverse functions

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Nov 4, 2016

#color(green)(dy/dx=2/sqrt(1-(2x+1)^2))#

Explanation:

Given:#" Determine "d/dx[sin^(-1)(2x+1)]#

#color(blue)("Preamble")#

Note that #sin^(-1)# has a particular meaning which has the alternative presentation of Arcsin. It is not connected to the form example of #sin^2# which is sin squared

So #sin^(-1)(2x+1) -> arcsin(2x+1)#

Note that #sin^(-1)(2x+1)# is another way of writing an angle and the sin of which gives the value #2x+1#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Answering the question")#

Set #theta =sin^(-1)(2x+1)#

Note that #sin(theta) = ("Opposite")/("Hypotenus") #

Set the value of the Hypotenuse to 1 giving:

#sin(theta) = ("Opposite")/("Hypotenuse") =(2x+1)/1#

Tony B

Using #sin(theta)=2x+1# and implicitly differentiating

#cos(theta)xxdy/dx =2#

#=>dy/dx=2/cos(theta)#.......................Equation(1)

But # "Hypotenuse "xxcos(theta) =" Adjacent "=x#

as the hypotenuse is 1 we have #cos(theta)=x#

So equation(1) becomes:

#dy/dx=2/x" .......................Equation"(1_a)#

But by Pythagoras #x^2=1^2-(2x+1)^2" "=>" "x=sqrt(1-(2x+1)^2)#

Thus #"Equation"(1_a)# becomes

#color(blue)(dy/dx=2/sqrt(1-(2x+1)^2))#