# Solve by Implicit Differentiation or Partial Differentiation?

## Homework Statement

I've got a question more with the structure of how this problem is presented:

If

x^(sin y) = y^(cos x)

Find
$$\frac{dx}{dy}(\frac{pi}{4},\frac{pi}{4})$$

## Homework Equations

We have been taught to solve by implicit differentiation when a problem looks something like this:

if x^3 + 2xy^2 +y^3 =1

But Ive seen some examples online of partial derivatives that present an expression with f (x,y)...but this problem looks different that that also.

The only thing in common is that there is an X and a Y variable.

## The Attempt at a Solution

Im unsure of what to look for to solve this...maybe its not even implicit differentiation.
I am going with setting the expression to 0 ie:

(sin y)
x^
------------- = 0
(cos x)
y^

and then solving implicitly...and then popping in the pi/4 into the equation.
Does this sound like a good idea? Or is my instructor trying to "trick" us ...lol
maybe its implicit differentiation WITH exponential/Logarithmic differentiation? Anyway...
Thanks for any clues.

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Implicit differentiation is simply using an implicit relationship between variables (such as that given by an equation instead of one variable being written as an explicit function of the other) to find the derivative. The form of the equation is irrelevant.
Note that if A(x) = B(x), then dA/dx = dB/dx.
The differentiation of each side of the equation with respect to y is a straightforward application of the chain rule, although you should first find the derivative of the related function f(x) = ax where a is a constant.

"The differentiation of each side of the equation with respect to y is a straightforward application of the chain rule, although you should first find the derivative of the related function f(x) = ax where a is a constant."

Are you saying all I need to do is differentiate each side?and it becomes a matter of simplifying the expression? (say in terms of sin()?) and then plugging in the pi/4?

Hurkyl
Staff Emeritus
Gold Member
maybe its implicit differentiation WITH exponential/Logarithmic differentiation?
That would be reasonable. You have to differentiate sometime to introduce derivatives, but you can do that whenever -- you can take a logarithm first if you like. (But be careful! Make sure things are positive....)

P.S. logarithmic differentiation is implicit differentiation of a specific type

P.P.S. What variable are you going to differentiate with respect to? Don't just pick x out of habit! Make an informed choice!

ah..its an inverse function differentiation right?

like:

xe^y = y - 1?

and can solve in that sense (we arent on that chapter yet...so Ill have to look at it)

Consider:

$x^{sin(y)} = y^{cos(x)} = \frac{ln(sin(y))}{ln(x)}=\frac{ln(cos(x))}{ln(y)}$

Now, as slider's pointed out, if A(x)=B(x) then dA(x)/dx=dB(x)/dx now if we say the left side (ln(sin(y))/ln(x)) is u and the right side is v consider the chain rule expansion:

$\frac{du}{dy}=\frac{dv}{dy}=\frac{dv}{dx}\frac{dx}{dy}$

I hope that helps you.

Hurkyl
Staff Emeritus
Gold Member
$x^{sin(y)} = y^{cos(x)} {\color{red} =} \frac{ln(sin(y))}{ln(x)}=\frac{ln(cos(x))}{ln(y)}$
That's not right, for two reasons.

(1) That red equality sign is completely wrong

(2) You took logarithms incorrectly

$x^{sin(y)} \rightarrow sin(y)ln(x), y^{cos(x)} \rightarrow cos(x)ln(y)$

In the region x,y (0,Pi/2)

Thanks for the clues maverick and everyone else! Im gonna try my attempt now and then get back with you see if Im close or not.

I am a little confused by what you meant maverick:
logarithmic differentiation is implicit differentiation of a specific type
Also why did you mention:
In the region x,y (0,Pi/2)
Is it supposed to be (0, pi/4)?

No, (0,Pi/2) because in order to use the logarithm identity $log(x^y)=ylog(x)$ y must be positive which, in this case y= cos(x) or sin(y) which are both positive in the region (0,Pi/2)

Are you saying all I need to do is differentiate each side?and it becomes a matter of simplifying the expression? (say in terms of sin()?) and then plugging in the pi/4?
That's it. However, note that since they are asking for dx/dy, it will be more straightforward to differentiate both sides with respect to y.

Thanks everyone...its starting to make sense...

No, (0,Pi/2) because in order to use the logarithm identity LaTeX Code: log(x^y)=ylog(x) y must be positive which, in this case y= cos(x) or sin(y) which are both positive in the region (0,Pi/2)
I see what you are saying maverick, BUT in regards to what Slider is saying:
Originally Posted by Liquid7800 View Post

Are you saying all I need to do is differentiate each side?and it becomes a matter of simplifying the expression? (say in terms of sin()?) and then plugging in the pi/4?
That's it. However, note that since they are asking for dx/dy, it will be more straightforward to differentiate both sides with respect to y.
I solve using the LOG identity and THEN plugin the (pi/4,pi/4) values? You were just stating what region the y-value is to use the Log identity?

Hurkyl
Staff Emeritus
Gold Member
I am a little confused by what you meant maverick:
(actually, it was my comment)

Logarithmic differentiation is the technique when faced with an expression defining y as a function of x such as this one (where x is a variable ranging over the positive reals)

y = x^x

you first take the logarithm of both sides

ln y = x ln x

(which is an expression implicitly relating y to x), and you use implicit differentiation to introduce y'

(1/y) y' = ln x + x (1/x)

and finish by solving for y':

y' = y (1 + ln x) = x^x (1 + ln x)

In fact, technically speaking, even differentiating the equation

y = sin x

to get

y' = cos x

is an example of implicit differentiation, albeit an utterly trivial one.