# Alternate way of differentiating x^y

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1. Jun 24, 2017

### randomgamernerd

1. The problem statement, all variables and given/known data: find the dy/dx of xy=a constant

2. Relevant equations: basic differentiation formulae

3. The attempt at a solution:

I know we can use logarithmic differentiation for differentiating x y..But can we differentiate it using chain rule and get answer as
yxy-1.dy/dx =0. ???

2. Jun 24, 2017

The answer is no. If $y=u^n$, (for $n$ being a constant), then $\frac{dy}{dx}=nu^{n-1} \frac{du}{dx }$. The operation you are attempting to do doesn't work.

3. Jun 24, 2017

### pasmith

The chain rule would give you $(y^x)' = (e^{(x \ln y)})' = (x \ln y)'y^x$ etc.

4. Jun 24, 2017

### Staff: Mentor

I think you meant $x^y$

5. Jun 24, 2017

### LCKurtz

Since $x^y$ is a power function of $x$ and an exponential function of $y$, which depends implicitly on $x$, you can't use either just the power rule or just the exponential rule for differentiation. But you can do it by the two-variable chain rule, if you have studied that. One form of it states that if $u$ and $v$ are functions of $x$ and you have a function $f(u,v)$ which you would like to differentiate with respect to $x$, can use$$\frac{df}{dx} = f_u\frac{du}{dx} + f_v\frac{dv}{dx}$$where the subscripts are partial derivatives. In your example $x^y = c$ think of $u=x$ and $v = y$ and $f(x,y) = x^y = c$. So you get$$f_x\frac{dx}{dx}+ f_y\frac{dy}{dx} = yx^{y-1}\cdot 1 + x^y \ln x \frac{dy}{dx} = 0$$If you solve that for $\frac{dy}{dx}$ and simplify it you will get the same answer as you get with logarithmic differentiation. Try it. You might find it interesting. Notice that in the process of doing that you use both the power and exponential rules for differentiation. Makes sense, no?

6. Jun 25, 2017

### randomgamernerd

Ok...Cool..

7. Jun 25, 2017

### randomgamernerd

Thanks Everyonee