# Implicit differentiation

#### A180S

##### New member
How 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.”

#### Dr.Peterson

##### Elite Member
You differentiate both sides, keeping the chain rule in mind, and then solve for dy/dx.

#### A180S

##### New member
I got ycos(xy). Is this correct?
u=sin(xy)
dy/dx=d/du[sin(xy)]•d/dx[x+y]
dy/dx=y cos(xy)•1
dy/dx=y cos(xy)

#### Subhotosh Khan

##### Super Moderator
Staff member
How 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.”
Let'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)

#### A180S

##### New member
Let'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)
dx/dy appears twice how do I solve for dy/dx = f(x,y)? Confused

#### HallsofIvy

##### Elite Member
What are you referring to? There is NO "dx/dy" in any of that.

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 $$\displaystyle \frac{dy}{dx}$$.

#### A180S

##### New member
What is dy/dx when solved?

#### Subhotosh Khan

##### Super Moderator
Staff member
What is dy/dx when solved?
You tell us.

#### Dr.Peterson

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

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

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

#### HallsofIvy

##### Elite Member
If you can't do algebra then you should not be attempting Calculus!
$$\displaystyle (y+ x\frac{dy}{dx})sin(x+ y)+ xycos(x+ y)(1+ \frac{dy}{dx})= 0$$.
$$\displaystyle y sin(x+ y)+ x sin(x+ y)\frac{dy}{dx}+ xy cos(x+y)+ xy cos(x+ y)\frac{dy}{dx}= 0$$
$$\displaystyle (x sin(x+ y)+ xy cos(x+ y))\frac{dy}{dx}+ y sin(x+ y)+ xy cos(x+ y)= 0$$
$$\displaystyle (x sin(x+ y)+ xy cos(x+ y))\frac{dy}{dx}= -y sin(x+ y)- xy cos(x+ y)$$
$$\displaystyle \frac{dy}{dx}= \frac{-y sin(x+ y)- xy cos(x+ y)}{x sin(x+ y)+ xy cos(x+ y)}$$.

#### A180S

##### New member
True I am working on my algebra, these problems really highlights any gaps in my knowledge (lots). I understand how the chain rule and product rule has been applied. I don’t understand how dy/dx (or A) was transposed/manipulated, what algebraic rules were used to do this? I researched over the weekend but can’t find any other examples showing a derivative being isolated this way.

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?

#### Subhotosh Khan

##### Super Moderator
Staff member
True I am working on my algebra, these problems really highlights any gaps in my knowledge (lots). I understand how the chain rule and product rule has been applied. I don’t understand how dy/dx (or A) was transposed/manipulated, what algebraic rules were used to do this? I researched over the weekend but can’t find any other examples showing a derivative being isolated this way.

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

#### Singleton

##### New member
maybe you write dy/dx as y'
in the above you just distribute into the parenthesis

#### JeffM

##### Elite Member
maybe you write dy/dx as y'
in the above you just distribute into the parenthesis
Yes. A derivative is just a function.

$$\displaystyle z = \dfrac{dy}{dx} \text { and } w = sin(x + y) \implies\\ \left (y + x * \dfrac{dy}{dx} \right ) * sin(x+y) = (y + xz)w = wy + wxzy =\\ ysin(x+y) + xsin(x+y) * \dfrac{dy}{dx}.$$

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.