Question

Find the derivative of f(x)=8(x^4) arctan (9x^3)

Answer 1

I think I'm making minor mistakes...

Answer 2

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

Answer 3

\[\text{ let } g=\arctan(9x^3) => \tan(g)=9x^3 =>g' \sec^2(g)=27x^2 \]
\[=> g'=\frac{72 x^2}{\sec^2(g)}\]
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...
\[f'=32x^3 \arctan(9x^3)+8x^4 \frac{72x^2}{8x^6+1}\]

Answer 4

oops 72 is suppose to be 27 lol

Answer 5

and 8 is suppose to be 81

Answer 6

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

Answer 7

frac{ }{ }