G'day,

If by "in-depth" you mean "to an Unco degree" and by "calculate" you mean "find"...

We have

\(\displaystyle y = 7\arcsin{x^2}\)

Divide both sides by 7:

\(\displaystyle \frac{y}{7} = \arcsin{x^2}\)

Take sin of both sides:

\(\displaystyle \sin{\frac{y}{7}} = x^2\)

Differentiate implicitly wrt x:

\(\displaystyle \frac{1}{7}\cos{\left(\frac{y}{7}\right)}\cdot \frac{dy}{dx} = 2x\)

\(\displaystyle \frac{dy}{dx} = \frac{14x}{\cos{\frac{y}{7}}\)

Now to put it terms of x only...

We had \(\displaystyle \sin{\frac{y}{7}} = x^2\) so can draw a right-angled triangle:

Code:

```
/| By Pythagoras, [sqrt(1-x^4)]^2 + (x^2)^2 = 1^2
/ |
1 / |
/ | x^2
/ |
/ y/7 |
/____________| y/7 is the angle, x^2 is opp
1 is the hyp, sqrt(1-x^4) is the adj
sqrt(1-x^4)
```

And from this \(\displaystyle \cos{\left(\frac{y}{7}\right)} = \frac{adj}{hyp} = \frac{\sqrt{1-x^4}}{1} = \sqrt{1-x^4}\)

So we have

\(\displaystyle \frac{dy}{dx} = \frac{14x}{\sqrt{1-x^4}}\)

You could use trig identities just as well.