## lgg23 4 years ago Find the derivative of f(x)=8(x^4) arctan (9x^3)

1. lgg23

I think I'm making minor mistakes...

2. anonymous

$8x ^{4}\times \tan^{-1} (9x ^{3}) = 32x ^{3}\times \tan^{-1} (9x ^{3}) + 8x ^{4}\times(27x ^{2}/(1+81x ^{6}))$

3. myininaya

$\text{ let } g=\arctan(9x^3) => \tan(g)=9x^3 =>g' \sec^2(g)=27x^2$ $=> g'=\frac{72 x^2}{\sec^2(g)}$ Now if tan(g)=9x^3/1=(opp/adj) Then assuming we have a right triangle the hyp is... $hyp=\sqrt{(9x^3)^2+1^2}=\sqrt{81x^6+1}$ So we have $g'=\frac{72x^2}{(\sqrt{81x^6+1})^2}=\frac{72x^2}{8x^6+1}$ so we have the derivative of f is... $f'=32x^3 \arctan(9x^3)+8x^4 \frac{72x^2}{8x^6+1}$

4. myininaya

oops 72 is suppose to be 27 lol

5. myininaya

and 8 is suppose to be 81

6. anonymous

yes :), @myiniaya one question, how i can write the line of division :/

7. myininaya

frac{ }{ }

8. myininaya

|dw:1328491764888:dw|

9. anonymous

$\frac{thank}{you}$ ok :)

10. anonymous

@igg23 , remember : $f(x)=\arctan(u)$ then $f \prime(x)=\frac{u \prime}{1+u ^{2}}$

11. myininaya

great job :)