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

Dec 25, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - 25 {x}^{2}}}$

#### Explanation:

$y = \arcsin \left(5 x\right) = {\sin}^{-} 1 \left(5 x\right)$

This implies that

$\sin y = 5 x$

So now we can differentiate $\sin y$

$f ' \left(\sin y\right) = \cos y$

Now we have $\frac{\mathrm{dx}}{\mathrm{dy}}$ but we want $\frac{\mathrm{dy}}{\mathrm{dx}}$. If we look at the two expressions, we can see that they are inverses of one another.

Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} y$

So now we have the derivative of our original equation, but we need to express $\cos y$ in terms of $x$.

We can do this using the Pythagorean rule:

${\sin}^{2} y + {\cos}^{2} y = 1$

${\cos}^{2} y = 1 - {\sin}^{2} y$

${\sin}^{2} y = {\left(\sin y\right)}^{2} = {\left(5 x\right)}^{2} = 25 {x}^{2}$

$\cos y = \sqrt{1 - 25 {x}^{2}}$

Dec 25, 2016

Use the Chain Rule . Please see the explanation.

#### Explanation:

The Chain Rule is:

$\frac{\mathrm{df} \left(g \left(x\right)\right)}{\mathrm{dx}} = \frac{\mathrm{df} \left(g\right)}{\mathrm{dg}} \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} \text{ [1]}$

Let $g \left(x\right) = 5 x$, then $f \left(g\right) = \arcsin \left(g\right) , \frac{\mathrm{df} \left(g\right)}{\mathrm{dg}} = \frac{1}{\sqrt{1 - {g}^{2}}} \mathmr{and} \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = 5$

$y ' = \frac{\mathrm{df} \left(g\right)}{\mathrm{dg}} \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} \text{ [2]}$

Substitute $\frac{1}{\sqrt{1 - {g}^{2}}}$ for $\frac{\mathrm{df} \left(g\right)}{\mathrm{dg}}$ in equation [2]:

$y ' = \frac{1}{\sqrt{1 - {g}^{2}}} \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} \text{ [3]}$

Substitute 5 for $\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}}$ in equation [3] but we can show it in the numerator:

$y ' = \frac{5}{\sqrt{1 - {g}^{2}}} \text{ [3]}$

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

$y ' = \frac{5}{\sqrt{1 - {\left(5 x\right)}^{2}}} \text{ [4]}$