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Math Help - finding derivatives and sketching inverse trig

  1. #1
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    finding derivatives and sketching inverse trig

    If f(x)=sin-1x+cos-1x find f '(x) and sketch the graph of y=f(x)

    If f(x) = cos-1(sinx) find f '(x) and sketch the graph of y=f(x) for -pi<or equalto x< or equal to pi

    thankyou =)
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  2. #2
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    Hello christina
    Quote Originally Posted by christina View Post
    If f(x)=sin-1x+cos-1x find f '(x) and sketch the graph of y=f(x)

    If f(x) = cos-1(sinx) find f '(x) and sketch the graph of y=f(x) for -pi<or equalto x< or equal to pi

    thankyou =)
    1 \frac{d}{dx}\Big(\arcsin(x)\Big)=\frac{1}{\sqrt{1-x^2}} and \frac{d}{dx}\Big(\arccos(x)\Big)=-\frac{1}{\sqrt{1-x^2}}, so f'(x) = 0, for all values of x.

    In fact f(x) = \frac{\pi}{2} (can you see why?), so the graph is a horizontal line at \frac{\pi}{2} for -1\le x \le 1

    2 Note that 0\le\arccos(x)\le\pi, for all valid values of x, and \frac{d}{dx}\Big(\arccos(x)\Big)=-\frac{1}{\sqrt{1-x^2}}. The derivative takes negative values since \arccos(x) decreases from \pi to 0 as x increases from -1 to +1.

    Thus, using the Chain Rule, if f(x) = \arccos(\sin x), f'(x) = -\frac{1}{\sqrt{1-\sin^2x}}\times \cos x=-1 while \sin x is increasing in the range -1 \le\sin x\le +1.

    Now \sin x increases from -1 to +1 as x increases from -\pi/2 to +\pi/2, so in this range f'(x) = -1. But for -\pi\le x\le -\pi/2,\, \sin x decreases from 0 to -1, and for \pi/2 \le x \le \pi,\, \sin x decreases from 1 to 0. Hence in these ranges f(x) is increasing, and therefore f'(x) = +1.

    The graph is as shown in the attached diagram.

    Grandad
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