Question

I will give medal and fan!!!
Help with calculus problems! (:

Answer 1

@lochana

Answer 2

use chain rule

Answer 3

\[\frac{ d }{ dx }(\sin 3x^2)=\cos 3x^2. \frac{ d }{ dx}(3x^2)\]

Answer 4

\[=6x \cos 3x^2\]

Answer 5

I go 6xcos(3x^2)

Answer 6

Oh okay cool. So what i got was correct. right?

Answer 7

yeah

Answer 8

Last one. The hardest to me,

Answer 9

its easy.. You use the chain rule

Answer 10

\[f'(x)=\frac{ 1 }{ x^3+e ^{4x} }.\frac{ d }{ dx }(x^3+e ^{4x})\]

Answer 11

tricky part is that you need to know
\[\frac{d(lnx)}{dx} = \frac{1}{x}\]

Answer 12

exactly @jango_IN_DTOWN is correct

Answer 13

\[=\frac{ 3x^2+4e^{4x} }{ x^3+e ^{4x} }\]

Answer 14

I suggest you read the theory portion of chain rule..

Answer 15

Its very useful.. You can do every problem yourself, I am quite sure, once you master it.:)

Answer 16

@arianna1453

Answer 17

yes. practice makes you perfect!

Answer 18

Thank you guys. I think I will practice that more. So the answer is going to be the 3x^2+4e^4x over x^3+e4x?

Answer 19

yeah

Answer 20

What you said above basically. It cannot be simplified more, correct?

Answer 21

no.. yoou dont have any common terms

Answer 22

So you take the outside and multiply it by the derivative of the inside?

Answer 23

do you want to learn about the chain rule?

Answer 24

Well i know that you take the outside, so ln and multiply it by the derivative of the inside the whole (x^3+e^4x)

Answer 25

Lets do the previous problem

Answer 26

f(x)=
\[\ln (x^3+e ^{4x})\]

Answer 27

Let
\[X=x^3+e ^{4x}\]

Answer 28

RIght,

Answer 29

then
f'(x)=
\[\frac{ d }{ dx}(\ln X)=\frac{ d }{ dX}(\ln X). \frac{ dX }{ dx }\]

Answer 30

\[=\frac{ 1 }{ X }.\frac{ d }{ dx }(x^3+e ^{4x})\]

Answer 31

okay and then you plug in (x^3+e^4x) in for x. Im following.

Answer 32

\[=\frac{ 1 }{ x^3 +e ^{4x}}.(3x^2+4e ^{4x})\]

Answer 33

for X

Answer 34

I understand it now!

Answer 35

yeah you let the entirely big term as X

Answer 36

Okay I have one more problem. I am going to try it myself and if I need help Ill tag you.

Answer 37

yeah sure

Answer 38

Okay so i did everything for this problem. starting with this.
\[(d/dx)(x) +2y (dy/dx) = (1)(y) + (x) (dy/dx)\]

Answer 39

@jango_IN_DTOWN

Answer 40

I got 1-y/x-2y.
So i think i messed up with the 2y part

Answer 41

first let me know if you know partial derivatives?

Answer 42

Like the xy part, correct?

Answer 43

wait see d(x+2y)=dx+2dy
and d(xy)=xdy+ydx
so from x+2y=xy we have dx+2dy=xdy+ydx
or, (2-x)dy=(y-1)dx
or. dy/dx=(y-1)/(2-x)

Answer 44

so where i had 2y (dy/dx) it was wrong there? it should just be 2(dy/dx)

Answer 45

I got the right side correct.

Answer 46

Anyways, thank you for your help today (:

Answer 47

ok.. bye.:)