## anonymous 5 years ago find the derivative of arcsin^3(x+3)

1. anonymous

Okay, here's the deal, we need to look at a general right triangle with hypot. 1 and vertical leg x

2. anonymous

This means that the horizontal leg has length $$\sqrt{1-x^2}$$.

3. anonymous

If we let $$y$$ be the angle between the horizontal leg and hypot., then $\sin y=opp/hyp=x$

4. anonymous

In other words, $$\arcsin x=y$$

5. anonymous

But then $$\cos(\arcsin x)=\cos y=adj/hyp=\sqrt{1-x^{2}}$$.

6. anonymous

Now, we are going to use the formula for derivative of inverse functions: If $$f$$ and $$g$$ are inverses $g^{\prime}=1/f^{\prime}(g)$

7. anonymous

Let $$f(x)=\sin x$$ and $$g(x)=\arcsin x$$. Plugging these in gives $(\arcsin x)^{\prime}=\frac{1}{\cos(\arcsin x)}=\frac{1}{\sqrt{1-x^{2}}}$

8. anonymous

using the chain rule and the definition of the derivative of arcsin I got 3[arcsinx]^2[1/(square root(1-x^2)]*[1/(square root(1-(x+3)^2)]

9. anonymous

no clue if this is correct

10. anonymous

Not quite, first thing is that arcsin x should be arcsin(x+3)

11. anonymous

Okay, your main problem is that when you take the chain rule, the first multiplicand (number in the product) is $$g^{\prime}(f(x))$$, that means that you need to put the whole $$f(x)$$ in there, not just $$x$$. So in the first chain rule, $$f(x)=arcsin(x+3)$$ meaning the first part of the product is $$3*(\arcsin(x+3))^{2}$$.

12. anonymous

ok got that thanks; what about the rest of it?

13. anonymous

In the second part, you need to replace x with x+3 and the last part is just (x+3)'=1