Question

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

Answer 1

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

Explanation:

`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

Use the Chain Rule . Please see the explanation.

Explanation:

The Chain Rule is:

`(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]"`