Question

#1. e^7x/x^6+1
#2. tan^-1 (3x)
#3. 7sin(x)sin^-1(x)
#4. arcsin^2(6x+6)
#5. 5cos(6 ln (x))
#6. 3sqrt4^x+7
#7. e^3x(x^2+6^x
#8. Find dy/dx in terms of x and y if ln y +y^3=6 ln x
#9. Find dy/dx in terms of x and y if 5xy+9x+y=8
#10. 2 ln (sinx) f"(x) ?

Answer 1

What is this?

Answer 2

this is derivatives. I posted a comment before this. I have most of the answers I wanted to go over them with someone on here.

Answer 3

Okay, so, for the first one, you have to find the derivative of\[f(x)=\frac{e^{7x}}{x^6}+1?\]

Answer 4

the bottom is x^6+1

Answer 5

I have to find derivatives for all 10 questions. but #10 i have to find the 2nd derivative. im doing something wrong but I cant figure out what it is.

Answer 6

\[f(x)=\frac{e^{7x}}{x^6+1}.\]Well, we need to know the quotient rule for this one. That is,\[f'(x)=\frac{(x^6+1)(e^{7x})'-(x^6+1)'(e^{7x})}{(x^6+1)^2}.\]That boils down to something like\[f'(x)=\frac{(x^6+1)(7e^{7x})-(6x)(e^{7x})}{(x^6+1)^2}.\]It's just a matter of simplifying things now...

Answer 7

Im good with the simplifying. I just needed to know how to get to that part.

Answer 8

10.\[f(x)=2\ln(\sin(x)).\]This requires we perform the chain rule to find the first derivative.\[\frac{df}{dx}=\frac{d}{dx}[2\ln(\sin(x))],\]\[\frac{df}{dx}=2\frac{d}{dx}[\ln(\sin(x))],\]\[\frac{df}{dx}=2\frac{1}{\sin(x)}\frac{d}{dx}[(\sin(x))],\]\[\frac{df}{dx}=2\frac{1}{\sin(x)}(\cos(x)),\]\[\frac{df}{dx}=2\cot(x).\]Can you find the derivative of the above expression?

Answer 9

-csc^2x ?

Answer 10

Very close:\[-2\cdot\csc^2(x).\]

Answer 11

ok. thanx. Lets go over the others??

Answer 12

Which other did you find difficult?

Answer 13

well, for # 2 i got sec^2(3^x) ??

Answer 14

Hmm, I wonder how you got that.

Answer 15

the derivastive of tan is sec^2 right?

Answer 16

2.\[f(x)=\tan^{-1}(3x).\]This one follows straight from the definition of the derivative of the inverse tangent function (if you don't feel like deriving it, that is).\[\frac{df}{dx}=\frac{d}{dx}[\tan^{-1}(3x)],\]\[\frac{df}{dx}=\frac{1}{1+(3x)^2}\frac{d}{dx}[3x],\]\[\frac{df}{dx}=\frac{3}{1+9x^2}.\]I think you confused these two:\[\tan(x)\neq\tan^{-1}(x).\]

Answer 17

For the record:\[\frac{d}{dx}[\tan^{-1}(x)]=\frac{1}{1+x^2}.\]

Answer 18

the answer i put in was (3) / (1+9x^2) but that is only part of the answer? right? Im not sure how to combine the two. does that make sense?

Answer 19

You mean, combine the terms in the denominator?

Answer 20

i mean take 1/1+x^2 and combine it with 3/1+9x^2 . Is that what im supposed to do to get my answer?

Answer 21

No, I was merely stating that\[\frac{d}{dx}[\tan^{-1}(x)]=\frac{1}{1+x^2}.\]However,\[\frac{d}{dx}[\tan^{-1}(3x)]=\frac{3}{1+9x^2},\]by applying the chain rule (see above).

Answer 22

ok. could you help me with #8 and 9? what does it mean "in terms of x and y?"

Answer 23

Questions 8 and 9 are asking you to perform implicit differentiation. I suppose you have heard of it before?

Answer 24

Say I give you an example, and you follow the train of thought to solve number 9?

Suppose I give you this:\[25x^2+y^2=109.\]Finding an implicit differential of this function, assuming y is the independent variable, goes as follows:\[\frac{d}{dx}[25x^2+y^2]=\frac{d}{dx}[109].\]Evaluating as usual should give you:\[{50x+2y\frac{dy}{dx}}=0.\]You then solve for dy/dx and obtain:\[\frac{dy}{dx}=-\frac{50x}{2y}=-25\frac{x}{y}.\]That's all there is to it.

Answer 25

cool!

Answer 26

I meant to say "assuming y is the dependent variable."

-.-

Let me know when you have given those a try.

Answer 27

ok...i have to jump off here for a few minutes. thanx for all your help.

Answer 28

No, you're welcome.