Question

Differentiate the following function.
f (x) = 7 x ^6 - 8 sin(x)

Answer 1

f'(x) = 42x^5 - 8Cosx

Answer 2

how did u get this? did you use the product rule

Answer 3

because there is a minus sign inbetween the terms, you just differentiate the terms seperately, no need for the product rule :)

Answer 4

if the problem had been:
\[7x^{6}\sin(x)\]
that calls for a product rule.

Answer 5

yea good point joemath..

Answer 6

is it product rule for this: \[8\sqrt{x}\sin x\]

Answer 7

showing the work, your problem is this:
\[\frac{d}{dx} (7x^{6}-8\sin(x)) = \frac{d}{dx}(7x^{6})-\frac{d}{dx}(8\sin(x)) = 7(6x^{5})-8(\cos(x))\]
\[=42x^{5}-8\cos(x)\]
Yes, it will be a product rule for that question.

Answer 8

product rule is:
\[(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)\]
\[f(x) = \sqrt{x} \Rightarrow f'(x) = \frac{1}{2\sqrt{x}}\]
\[g(x)=\sin(x)\Rightarrow g'(x)=\cos(x)\]
Plugging into the product rule we obtain:
\[(\sqrt{x}\sin(x))'=\frac{1}{2\sqrt{x}}(\sin(x))+\sqrt{x}\cos(x)\]
Multiply everything by 8 to get the correct answer :)

Answer 9

Why do I multiply everything by 8....Why wasnt the 8 left in the equation

Answer 10

Im using this rule:
\[\frac{d}{dx}(cf(x)) = c(\frac{d}{dx}f(x))\]
You can pull out constants when taking a derivative. So:
\[(8\sqrt{x}\sin(x))' = 8(\sqrt{x}\sin(x))'\]
I left out the 8, got an answer, and now i multiply by 8 make up for what i did earlier.

Answer 11

so is it \[8\sin x+8\sqrt{x}\cos x\]

Answer 12

sorry i forgot the 4/sqrt(x) in the front

Answer 13

the answer I got leaving out the 8 was:
\[(\sqrt{x}\sin(x))' = \frac{1}{2\sqrt{x}}\sin(x)+\sqrt{x}\cos(x)\]
So putting the 8 back in would give:
\[8(\sqrt{x}\sin(x))'=\frac{4}{\sqrt{x}}\sin(x)+8\sqrt{x}\cos(x)\]
Yeah you got it :)

Answer 14

g (t) = 6 t ^3 sin(t)

for this one is the answer this:

\[(6t^3)(\cos t)+(\sin t)(18t^2)\]

Answer 15

thats correct :)

Answer 16

its telling me its wrong tho

Answer 17

hmm...i guess maybe it wants the answer simplified? there is some factoring we can do:
\[6t^{3}\cos(t)+18t^{2}\sin(t) = 6t^{2}(tcos(t)+3\sin(t))\]

Answer 18

You answer is correct though, dont think you did something wrong.

Answer 19

your*

Answer 20

yup...it wanted it simplified

Answer 21

thats dumb >.> at least you got it though :)

Answer 22

ok one more!

y = csc(θ) (θ + cot(θ))

Answer 23

Lets see:
\[f(\theta) = \csc(\theta) \Rightarrow f'(\theta)=-\csc(\theta)\cot(\theta)\]
\[g(\theta)=\theta+\cot(\theta)\Rightarrow g'(\theta)=1-\csc^{2}(\theta)=cot^{2}(\theta)\]
So putting this all in the product rule we have:
\[(\csc(\theta)(\theta+\cot(\theta)))'=(-\csc(\theta)\cot(\theta))(\theta+\cot(\theta))+\csc(\theta)\cot^{2}(\theta)\]

Answer 24

Lets simplify... one sec

Answer 25

\[=-\theta \csc(\theta)\cot(\theta)-\csc(\theta)\cot^{2}(\theta)+\csc(\theta)\cot^{2}(\theta)=-\theta \csc(\theta)\cot(\theta)\]

Answer 26

I hate trig functions =/

Answer 27

i cant see the answer

Answer 28

\[-\theta \csc(\theta)\cot(\theta)\]

Answer 29

can you see it?

Answer 30

yes i can thanks

Answer 31

it say its wrong...this is frustrating

Answer 32

ugh, let me look over my work to make sure i didnt do anything wrong first...

Answer 33

I was a little hasty with my trig identity, uploading a pic of the solution in a sec

Answer 34

ok thanks

Answer 35

thanks alot it worked!

Answer 36

Are you still there?