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Math Help - Trigonometry Function Derivative

  1. #1
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    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}}.


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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 (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
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    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.
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  4. #4
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    Quote Originally Posted by Air View Post
    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.
    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..
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  5. #5
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    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}
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