- Thread starter A180S
- Start date

- Joined
- Nov 12, 2017

- Messages
- 8,036

If you're having trouble doing that, please show your work so we can see what help you need.

- Joined
- Jun 18, 2007

- Messages
- 21,305

Let's do this step-by-stepHow do you solve this?

“If (x+y)sin(xy)=1, use implicit differentiation to determine dy/dx, expressing your answer in the form dy/dx = f(x,y), where f(x,y) is an expression that involves terms in both x and y.”

d/dx (x*y) = x * dy/dx + y ..............(1)............ product rule

d/dx[sin(xy)] = [cos(xy)] * d/dx (x*y) = [x * dy/dx + y] * cos(x*y) ..........................(2) ..........................chain-rule

d/dx (x + y) = 1 + dy/dx.....................................(3)

d/dx [(x+y) * sin(x*y)] = [d/dx (x+y)] * sin(xy) + (x + y) * d/dx [sin(xy)] ..............................................(4) ............... product rule

Now put (1), (2) and (3) into (4)

dx/dy appears twice how do I solve for dy/dx = f(x,y)? ConfusedLet's do this step-by-step

d/dx (x*y) = x * dy/dx + y ..............(1)............ product rule

d/dx[sin(xy)] = [cos(xy)] * d/dx (x*y) = [x * dy/dx + y] * cos(x*y) ..........................(2) ..........................chain-rule

d/dx (x + y) = 1 + dy/dx.....................................(3)

d/dx [(x+y) * sin(x*y)] = [d/dx (x+y)] * sin(xy) + (x + y) * d/dx [sin(xy)] ..............................................(4) ............... product rule

Now put (1), (2) and (3) into (4)

- Joined
- Jan 27, 2012

- Messages
- 6,293

To determine dy/dx from \(\displaystyle xy sin(x+ y)= 1\) start with

\(\displaystyle d/dx(xy sin(x+ y))= (xy)_x sin(x+ y)+ xy (sin(x+y))_x= 0\).

\(\displaystyle (xy)_y= y+ x\frac{dy}{dx}\) so the first term is \(\displaystyle (y+ x\frac{dy}{dx})sin(x+ y)\)

\(\displaystyle (sin(x+ y))_x= cos(x+ y)(1+\frac{dy}{dx}\) so the second term is \(\displaystyle xycos(x+y)(1+ \frac{dy}{dx})\).

Putting those together, \(\displaystyle (y+ x\frac{dy}{dx})sin(x+ y)+ xycos(x+ y)(1+ \frac{dy}{dx})= 0\).

Solve that for \(\displaystyle \frac{dy}{dx}\).

- Joined
- Jun 18, 2007

- Messages
- 21,305

You tell us.What is dy/dx when solved?

After we review your work/answer, if needed, we will guide you to the correct answer.

- Joined
- Nov 12, 2017

- Messages
- 8,036

One thing that sometimes helps is to replace dy/dx with some variable, maybe A to make it stand out:What is dy/dx when solved?

\(\displaystyle (y+ xA)\sin(x+ y)+ xy\cos(x+ y)(1+ A)= 0\)

Then just solve for A. You'll be distributing, collecting terms with A, and so on.

- Joined
- Jan 27, 2012

- Messages
- 6,293

\(\displaystyle (y+ x\frac{dy}{dx})sin(x+ y)+ xycos(x+ y)(1+ \frac{dy}{dx})= 0\).

\(\displaystyle y sin(x+ y)+ x sin(x+ y)\frac{dy}{dx}+ xy cos(x+y)+ xy cos(x+ y)\frac{dy}{dx}= 0\)

\(\displaystyle (x sin(x+ y)+ xy cos(x+ y))\frac{dy}{dx}+ y sin(x+ y)+ xy cos(x+ y)= 0\)

\(\displaystyle (x sin(x+ y)+ xy cos(x+ y))\frac{dy}{dx}= -y sin(x+ y)- xy cos(x+ y)\)

\(\displaystyle \frac{dy}{dx}= \frac{-y sin(x+ y)- xy cos(x+ y)}{x sin(x+ y)+ xy cos(x+ y)}\).

1) How does (y+xdy/dx)sin(x+y) become ysin(x+y)+xsin(x+y)dy/dx?

2) How does xycos(x+y)(1+dy/dx) become

xycos(x+y)+xycos(x+y)dy/dx?

- Joined
- Jun 18, 2007

- Messages
- 21,305

1) How does (y + x*dy/dx) * sin(x+y) become y*sin(x+y) + x*sin(x+y)*dy/dx

1) How does (y+xdy/dx)sin(x+y) become ysin(x+y)+xsin(x+y)dy/dx?

2) How does xycos(x+y)(1+dy/dx) become

xycos(x+y)+xycos(x+y)dy/dx?

In these cases,

treat dy/dx as an "unknown number" (or as a variable).

If I say

(A + B*C) * D = A * D + B * C * D ............................... would you agree with that?

We call it process of distribution - you probably learned it algebra as the "FOIL" method.

Same process (FOIL) for your Q2.

These are very basic rules of algebra.

Yes. A derivative is just a function.maybe you write dy/dx as y'

in the above you just distribute into the parenthesis

\(\displaystyle z = \dfrac{dy}{dx} \text { and } w = sin(x + y) \implies\\

\left (y + x * \dfrac{dy}{dx} \right ) * sin(x+y) = (y + xz)w = wy + wxzy =\\

ysin(x+y) + xsin(x+y) * \dfrac{dy}{dx}.\)

However, to remind us that the function is a derivative we use y' instead of z. Sometimes Leibniz notation is helpful; sometimes the pseudo_Newton notation is helpful.