# Derivative of Arcsine and Arctangent?

#### legacyofpiracy

##### Junior Member
I was wondering if anyone could offer a somewhat in-depth explanation of how you would go about calculating the derivative of the arcsine/tangent/secant?

Particularily these problem..the i can just apply the method to the other ones

If F(x)=7arcsine(x^2)
F'(x)= ?

and

if F(x)=2arctan(8x^7)
F'(x)= ?

Thanks alot.

#### Unco

##### Senior Member
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.

#### legacyofpiracy

##### Junior Member
Thanks you so much, this really helped