## nickymarden Group Title derive the funtion. 2 years ago 2 years ago

1. nickymarden Group Title

2. satellite73 Group Title

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

3. nickymarden Group Title

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. satellite73 Group Title

$(\frac{f}{g})'=\frac{gf'-f'g}{g^2}$ 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. satellite73 Group Title

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. satellite73 Group Title

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. nickymarden Group Title

yeees. :)

8. satellite73 Group Title

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. satellite73 Group Title

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. satellite73 Group Title

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. satellite73 Group Title

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. nickymarden Group Title

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

13. satellite73 Group Title

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. satellite73 Group Title

one is more than enough yw

15. satellite73 Group Title

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

16. nickymarden Group Title

but he wants it like that lol thank you :)