## anonymous 4 years ago derive the funtion.

1. anonymous

2. anonymous

$f(x)=\frac{\sin^2(x^2+5)}{\sqrt{x^2+1}}$? like that? will be an algebra nightmare your math teacher must hate you

3. anonymous

haha, that's right. I was supposed t do it in class, but i got to the middle and got fed up. Can you help me? please..

4. anonymous

$(\frac{f}{g})'=\frac{gf'-f'g}{g^2}$ with $f(x)=\sin^2(x^2+5),g(x)=\sqrt{x^2+1}$ we need both $$f'$$ and $$g'$$ and both require the chain rule

5. anonymous

so first we note that $$f$$ is actually a composition of three functions, $$x^2+5$$ , sine and the squaring function so good $f'(x)=2\sin(x^2+5)\times \cos(x^2+5)\times 2x=4x\sin(x^2+5)\cos(x^2+5)$

6. anonymous

and also by the chain rule $g'(x)=\frac{2x}{2\sqrt{x^2+1}}=\frac{x}{\sqrt{x^2+1}}$ so far so good?

7. anonymous

yeees. :)

8. anonymous

now we have to put this mess together for $(\frac{f}{g})'=\frac{gf'-f'g}{g^2}$ the denominator is easy enough, it is just $$x^2+1$$ but the numerator is going to be a colossal mess i guess we can just write it down

9. anonymous

numerator is $\sqrt{x^2+1}\times 4x\sin(x^2+5)\cos(x^2+5)-\sin^2(x^2+5)\times \frac{x}{\sqrt{x^2+1}}$

10. anonymous

and so your whole mess is $\frac{\sqrt{x^2+1}\times 4x\sin(x^2+5)\cos(x^2+5)-\sin^2(x^2+5)\times \frac{x}{\sqrt{x^2+1}}}{x^2+1}$

11. anonymous

now you can get rid of the compound fraction by multiplying top and bottom by $$\sqrt{x^2+1}$$ to get $\frac{(x^2+1)\times 4x\sin(x^2+5)\cos(x^2+5)-\sin^2(x^2+5)\times x}{(x^2+1)\sqrt{x^2+1}}$ and maybe you can even clean this mess up with some more algebra by multiplying out in the numerator or something, but my algebra is not that good

12. anonymous

omg, if i could give you a million medals i would.

13. anonymous

there is also an annoying trig identity you can use since $2\sin(x)\cos(x)=\sin(2x)$ so if you want to be really clever you can replace $4x\sin(x^2+5)\cos(x^2+5)$ by $2x\sin(2x^2+10)$ but i wouldn't

14. anonymous

one is more than enough yw

15. anonymous

oh also you can write the denominator as $$(x^2+1)^{\frac{3}{2}}$$ but that strikes me as unnecessary also

16. anonymous

but he wants it like that lol thank you :)