Question

Compute dy/dx: x^4y^3+tany=sinx

Answer 1

differentiate term by term

Answer 2

what is the derivative of x^4y^3 ?

Answer 3

(use the product rule and the chain rule for y - i.e. to multiply times dy/dx the derivative of y)

Answer 4

I thought she was wanting us to use the implicit differentiation

Answer 5

yes, we are going to

Answer 6

I mean there is no way to go otherwise...

Answer 7

k, do you know how to compute the derivative of x^4y^3 ?

Answer 8

4x^3(3y^2)?

Answer 9

there is a thing called a product rule for the derivative of a product of the functions f(x) times g(x).
\(\large\color{black}{ \displaystyle f'(x)g(x)+f(x)g'(x) }\)

Answer 10

\(\large\color{black}{ \displaystyle x^4\times y^3 ~~~_{\Huge \Longrightarrow}^{\rm becomes}~~~\color{red}{4x^3}\times y^3~+ ~x^4 \times \color{red}{3y^2 \times \frac{dy}{dx}} }\)

in red are the derivatives.

Answer 11

see how I am differentiating the product of these 2 functions?

Answer 12

Yes

Answer 13

ok, can you differentiate the second term?
(( derivative of tan(y), and don't forget that the derivative of y will be always multiplied times dy/dx ))

Answer 14

always **be** multiplied...

Answer 15

at first, what is the derivative of tan(x) ?

Answer 16

sec^2x

Answer 17

yes derivative of tan(x) = sec^2(x).
and derivative of tan(y) = sec^2(y) \(\small\color{black}{ \displaystyle \times }\) (dy/dx)

Answer 18

see how the derivative of y works ?
(multiplying the derivative of y times dy/dx every time)

Answer 19

so, the derivative of tan(y) is what (again) ?

Answer 20

sec^2(y) haha

Answer 21

Yes

Answer 22

you skipped dy/dx

Answer 23

(dy/dx) * sec^2(y)

Answer 24

the derivative of y is same as derivative of x, BUT derivative of y is multiplied times dy/dx

Answer 25

and what is the derivative of sin(x) ?

Answer 26

cos(x)

Answer 27

yes, so putting the terms together now...

Answer 28

\(\large\color{black}{ \displaystyle ~~~~~~~~~~x^4y^3~~~~~~~~~~~~~~~~+\tan y~~~~~~~~~~=\sin x }\)

\(\large\color{black}{ \displaystyle 4x^3y^3~+ ~x^4 3y^2 \color{red}{\frac{dy}{dx}}~~+\sec^2y\color{red}{\frac{dy}{dx}}~~~=\cos x }\)

Answer 29

I put the derivatives underneath.... see ?

Answer 30

Now, all you need to do is to isolate the dy/dx on one side.

Answer 31

Yes

Answer 32

can you isolate (algebraically solve for) dy/dx ?

Answer 33

I'm not sure, since dy/dx isn't in all the terms. I feel like I need to thoug

Answer 34

though

Answer 35

\(\large\color{slate}{\large\color{black}{ \displaystyle 4x^3y^3~+ ~x^4 3y^2 \color{red}{\frac{dy}{dx}}~~+\sec^2y\color{red}{\frac{dy}{dx}}~~~=\cos x }}\)

\(\large\color{slate}{\large\color{black}{ \displaystyle x^4 3y^2 \color{red}{\frac{dy}{dx}}~~+\sec^2y\color{red}{\frac{dy}{dx}}~~~=\cos x -4x^3y^3 }}\)

\(\large\color{slate}{\large\color{black}{ \displaystyle \color{red}{\frac{dy}{dx}}\left( x^4 3y^2+\sec^2y\right)=\cos x -4x^3y^3 }}\)

\(\large\color{slate}{\large\color{black}{ \displaystyle \color{red}{\frac{dy}{dx}}=\frac{\cos x -4x^3y^3}{x^4 3y^2+\sec^2y} }}\)

Answer 36

that is all....

Answer 37

Thank you

Answer 38

yw