finding derivatives and sketching inverse trig

**christina** · November 4th 2009, 02:53 AM

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 =)

**Grandad** · November 4th 2009, 05:22 AM

Hello christina

Originally Posted by christina

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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