Question

What is the derivative of t^4/(t^8+1)^(1/2)

Answer 1

I used product rule.
\[f \prime(t)=t ^{4}(t ^{8}+1)^{-1/2}\]=

\[4t ^{3}(t ^{8}+1)^{-1/2}+-.5(t ^{8}+1)^{-3/2}(8t ^{7})(t ^{4})\]

Find my error please, I am pretty sure it is incorrect.

Answer 2

\[\frac{d}{dx}(t^4 \cdot (t^8+1)^{-1/2})\]Product rule: f'g + g'f where \(f=t^4\) and \(g=(t^8+1)^{-1/2}\)\[=4t^3(t^8+1)^{-1/2} -\dfrac{1}{2}(t^8+1)^{-3/2}\cdot (8t^7)\cdot t^4\]\[=\frac{4t^3}{(t^8+1)^{1/2}} -\frac{(4t^7)(t^4)}{(t^8+1)^{3/2}}\]\[=\frac{4t^3(t^8+1)}{(t^8+1)^{3/2}}-\frac{4t^{11}}{(t^8+1)^{3/2}}\]\[=\frac{4t^3(t^8+1)-4t^{11}}{(t^8+1)^{3/2}}\] Now simplify this.

Answer 3

thank you, I need help simplifying something
\[\sqrt{\frac{ 4t ^{3} }{ (t ^{8}+1)^{3/2} }^{2}-\frac{ 4t ^{7} }{ (t ^{8}+1)^{3/2} }^{2}}\]

Answer 4

\[=\frac{4t^3(t^8+1)-4t^{11}}{(t^8+1)^{3/2}}\]\[=\frac{4t^{11}+4t^3-4t^{11}}{(t^8+1)^{3/2}}\]\[=\boxed{\dfrac{4t^3}{(t^8+1)^{3/2}}} \]

Answer 5

If you were referring to the same problem with that complicated question.

Answer 6

I got that, I need help with above simplification now.

Answer 7

\[\large \sqrt{\left(\frac{ 4t ^{3} }{ (t ^{8}+1)^{3/2} }\right)^{2}-\left(\frac{ 4t ^{7} }{ (t ^{8}+1)^{3/2} }\right)^{2}}\] Is it like this?

Answer 8

I got
\[\frac{ 16t ^{6} }{(t ^{8}+1)^{2} }\]

Answer 9

no, it is plus sign. sorry i wrote it wrong

Answer 10

\[\large \sqrt{\left(\frac{ 4t ^{3} }{ (t ^{8}+1)^{3/2} }\right)^{2}\color{red}+\left(\frac{ 4t ^{7} }{ (t ^{8}+1)^{3/2} }\right)^{2}}\]

Answer 11

that?

Answer 12

yes

Answer 13

@dan815 how would you go about solving this other problem, scroll down.

Answer 14

i see nothing wrong with ur very first comment