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Thread: Trigonometry Function Derivative

  1. #1
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    Trigonometry Function Derivative

    Q: $\displaystyle f(x) = \mathrm{arcsin} x, \ -1 \le x \le 1$.

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

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


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    I've done part (a) but struggling with part (b). May I have some help please.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    The equation of the tangent to the curve at a point $\displaystyle (a,\,f(a))$ is $\displaystyle y=f(a)+(x-a)\cdot f'(a)$. In your case, it becomes $\displaystyle y=f\left(\frac{1}{\sqrt{2}}\right)+\left(x-\frac{1}{\sqrt{2}}\right)\cdot f'\left(\frac{1}{\sqrt{2}}\right)=\ldots$
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    But...I still don't get it very well. I understand that the coordinates to consider is $\displaystyle \left( \frac{1}{\sqrt{2}}, \frac{\pi}{2} \right)$. Can someone work out the gradient? I think I might have got that wrong.
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    Quote Originally Posted by Air View Post
    Q: $\displaystyle f(x) = \mathrm{arcsin} x, \ -1 \le x \le 1$.

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

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


    __________________
    I've done part (a) but struggling with part (b). May I have some help please.
    The equation of the tangent to a curve whose equation is f(x) is :

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

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

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


    --> tangent :

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

    etc..
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  5. #5
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    Another way of getting the gradient:

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

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

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