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 =)
Hello christina1 $\displaystyle \frac{d}{dx}\Big(\arcsin(x)\Big)=\frac{1}{\sqrt{1-x^2}}$ and $\displaystyle \frac{d}{dx}\Big(\arccos(x)\Big)=-\frac{1}{\sqrt{1-x^2}}$, so $\displaystyle f'(x) = 0$, for all values of $\displaystyle x$.
In fact $\displaystyle f(x) = \frac{\pi}{2}$ (can you see why?), so the graph is a horizontal line at $\displaystyle \frac{\pi}{2}$ for $\displaystyle -1\le x \le 1$
2 Note that $\displaystyle 0\le\arccos(x)\le\pi$, for all valid values of $\displaystyle x$, and $\displaystyle \frac{d}{dx}\Big(\arccos(x)\Big)=-\frac{1}{\sqrt{1-x^2}}$. The derivative takes negative values since $\displaystyle \arccos(x)$ decreases from $\displaystyle \pi$ to $\displaystyle 0$ as $\displaystyle x$ increases from $\displaystyle -1$ to $\displaystyle +1$.
Thus, using the Chain Rule, if $\displaystyle f(x) = \arccos(\sin x), f'(x) = -\frac{1}{\sqrt{1-\sin^2x}}\times \cos x=-1$ while $\displaystyle \sin x$ is increasing in the range $\displaystyle -1 \le\sin x\le +1$.
Now $\displaystyle \sin x$ increases from $\displaystyle -1$ to $\displaystyle +1$ as $\displaystyle x$ increases from $\displaystyle -\pi/2$ to $\displaystyle +\pi/2$, so in this range $\displaystyle f'(x) = -1$. But for $\displaystyle -\pi\le x\le -\pi/2,\, \sin x$ decreases from $\displaystyle 0$ to $\displaystyle -1$, and for $\displaystyle \pi/2 \le x \le \pi,\, \sin x$ decreases from $\displaystyle 1$ to $\displaystyle 0$. Hence in these ranges $\displaystyle f(x)$ is increasing, and therefore $\displaystyle f'(x) = +1$.
The graph is as shown in the attached diagram.
Grandad