Question

Find dy/dx of e^(2x)=sin(x+4y)

Answer 1

well, the derivative of both sides will need the chain rule, and the derivative of 4y (which will be in the chain rule) will have a chain rule of (multiplying by) dy/dx.

Answer 2

then as you differentiate, isolate the dy/dx by itself to find the answer.

Answer 3

lets take it piece by piece.
What is the derivative of \(\large\color{black}{ \displaystyle e^{2x} }\) ?

Answer 4

2e^(2x)

Answer 5

yes.

Answer 6

What is the derivative of sin(x) ?

Answer 7

cos(x)

Answer 8

yes, but when it is sin(x+4y), (when the inner argument is x+4y) we will need the chain rule.

Answer 9

what is the derivative of x+4y (derivatiev dy/dx) ?

Answer 10

1+4dy/dx? I think this is where I am messing up

Answer 11

yes

Answer 12

So, sin(x+4y) comes as follows,

\(\large\color{black}{ \displaystyle \cos(x+4y)\times\left[1+4\frac{dy}{dx}\right] }\)

Answer 13

saying we differnetiate the entire thing as if it is just sin(x), and then applied the chain rule

Answer 14

cos(x+4y)*(1+4dy/dx)! I didn't use parenthesis for the )1+4dy/dx) I think that was my error

Answer 15

So, from both sides,
\(\large\color{black}{ \displaystyle e^{2x}=\sin(x+4y) }\)

differentiating,
\(\large\color{black}{ \displaystyle 2e^{2x}=\cos(x+4y)\times\left[1+4\frac{dy}{dx}\right] }\)

Answer 16

yes, parenthesis are very important

Answer 17

See how we differentiated the function ?

Answer 18

Yep! Can you work me through setting the side equal to dy/dx just for practice?

Answer 19

sure, you will need to expand the right side at first.

Answer 20

can you do that for me?

Answer 21

cos(x)+cos(4y)*(1+4dy/dx)? Or no? Rusty at math sorry I know this should be the simple part!

Answer 22

where does cos(x) come from?

Answer 23

\(\large\color{black}{ \displaystyle 2e^{2x}=\cos(x+4y)\times\left[1+4\frac{dy}{dx}\right] }\)

into,

\(\large\color{black}{ \displaystyle 2e^{2x}=\cos(x+4y)\times\left[1\right] +\cos(x+4y)\times\left[4\frac{dy}{dx}\right]}\)

Answer 24

\[A(b+C)=Ab+AC\]

Answer 25

(same here)

Answer 26

\(\large\color{black}{ \displaystyle 2e^{2x}=\cos(x+4y)\times\left[1\right] +\cos(x+4y)\times\left[4\frac{dy}{dx}\right]}\)

\(\large\color{black}{ \displaystyle 2e^{2x}=\cos(x+4y)+4\cos(x+4y)\times\left[\frac{dy}{dx}\right]}\)

Answer 27

then solve for dy/dx.... can you go on?

Answer 28

Just subtract them over right? Hahah

Answer 29

yes, subtract the cos(x+4y) from both sides, and then divide both sides by 4cos(x+4y)

Answer 30

Thank you so much!

Answer 31

sure, yw!