# Trigonometry Function Derivative

• May 8th 2008, 05:57 AM
Simplicity
Trigonometry Function Derivative
Q: $f(x) = \mathrm{arcsin} x, \ -1 \le x \le 1$.

(a)Evaluate $f \left( - \frac{1}{2} \right)$.

(b) Find an equation of the tangent to the curve with equation $y=f(x)$ at the point where $x = \frac{1}{\sqrt{2}}$.

__________________
I've done part (a) but struggling with part (b). May I have some help please. (Smile)
• May 8th 2008, 06:04 AM
flyingsquirrel
Hi

The equation of the tangent to the curve at a point $(a,\,f(a))$ is $y=f(a)+(x-a)\cdot f'(a)$. In your case, it becomes $y=f\left(\frac{1}{\sqrt{2}}\right)+\left(x-\frac{1}{\sqrt{2}}\right)\cdot f'\left(\frac{1}{\sqrt{2}}\right)=\ldots$
• May 8th 2008, 06:08 AM
Simplicity
But...I still don't get it very well. I understand that the coordinates to consider is $\left( \frac{1}{\sqrt{2}}, \frac{\pi}{2} \right)$. Can someone work out the gradient? I think I might have got that wrong.
• May 8th 2008, 06:10 AM
Moo
Yop

Quote:

Originally Posted by Air
Q: $f(x) = \mathrm{arcsin} x, \ -1 \le x \le 1$.

(a)Evaluate $f \left( - \frac{1}{2} \right)$.

(b) Find an equation of the tangent to the curve with equation $y=f(x)$ at the point where $x = \frac{1}{\sqrt{2}}$.

__________________
I've done part (a) but struggling with part (b). May I have some help please. (Smile)

The equation of the tangent to a curve whose equation is f(x) is :

$y=f'(a)(x-a)+f(a)$

Here, $a=\frac{1}{\sqrt{2}}$

$f(x)=\arcsin(x) \Longrightarrow f'(x)=\frac{1}{\sqrt{1-x^2}}$

--> tangent :

$y=\frac{1}{\sqrt{1-(\frac{1}{\sqrt{2}})^2}} \cdot (x-\frac{1}{\sqrt{2}})+f(\frac{1}{\sqrt{2}})$

etc..
• May 8th 2008, 06:20 AM
Gusbob
Another way of getting the gradient:

Gradient of Function y = f'(x) is $\frac{dy}{dx}$

$y = arcsin(x)$
$x = sin(y)$
$\frac{dx}{dy} = cos(y)$
$\frac{dy}{dx} = sec(y)$
but y = arcsin(x)

$\frac{dy}{dx} = sec(arcsin(x))$
at $x = \frac{1}{\sqrt2}$
$\frac{dy}{dx} = sec(\frac{\pi}{4})$
= $\sqrt{2}$