Alternate way of differentiating x^y

[randomgamernerd]

Homework Statement

: [/B]find the dy/dx of x^y=a constant

Homework Equations

: basic differentiation formulae[/B]

The Attempt at a Solution

:[/B]

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

 

[Charles Link]

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.

 

[pasmith]

Homework Statement

: [/B]find the dy/dx of x^y=a constant

Homework Equations

: basic differentiation formulae[/B]

The Attempt at a Solution

:[/B]

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

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

 

[Chestermiller]

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

I think you meant $x^y$

 

[LCKurtz]

Homework Statement

: [/B]find the dy/dx of x^y=a constant

Homework Equations

: basic differentiation formulae[/B]

The Attempt at a Solution

:[/B]

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

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?

 

[randomgamernerd]

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?

Ok...Cool..

 

[randomgamernerd]

Thanks Everyonee