How do you find the derivative of #y = arcsin(5x)#?

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Answer 1 · 2016-12-25T22:41:31

#dy/dx=1/sqrt(1-25x^2)#

#y=arcsin(5x)=sin^-1(5x)#

This implies that

#siny=5x#

So now we can differentiate #siny#

#f'(siny)=cosy#

Now we have #dx/dy# but we want #dy/dx#. If we look at the two expressions, we can see that they are inverses of one another.

Therefore, #dy/dx=1/cosy#

So now we have the derivative of our original equation, but we need to express #cosy# in terms of #x#.

We can do this using the Pythagorean rule:

#sin^2y+cos^2y=1#

#cos^2y=1-sin^2y#

#sin^2y=(siny)^2=(5x)^2=25x^2#

#cosy=sqrt(1-25x^2)#

Answer 2 · 2016-12-25T23:27:43

Use the Chain Rule . Please see the explanation.

#(df(g(x)))/dx = (df(g))/(dg)(dg(x))/dx" [1]"#

Let #g(x) = 5x#, then #f(g) = arcsin(g), (df(g))/(dg) = 1/sqrt(1 - g^2) and (dg(x))/dx = 5#

#y' = (df(g))/(dg)(dg(x))/dx" [2]"#

Substitute #1/sqrt(1 - g^2)# for #(df(g))/(dg)# in equation [2]:

#y' = 1/sqrt(1 - g^2)(dg(x))/dx" [3]"#

Substitute 5 for #(dg(x))/(dx)# in equation [3] but we can show it in the numerator:

#y' = 5/sqrt(1 - g^2)" [3]"#

Reverse the substitution for g by substituting 5x for g in equation [3]:

#y' = 5/sqrt(1 - (5x)^2)" [4]"#