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What is the partial derivative of $x$ with respect to $x^2$?

$$\frac\partial{\partial x^2}x$$It is zero? What is the derivative in this case? Can you provide an explanation.

lioness99a
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Gaurav Gupta
  • 139

3 Answers3

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By a chain rule we have $$\frac{df}{dx}=\frac{df}{d(x^2)}\cdot \frac{d(x^2)}{dx}$$

If $f(x)=x$, then we arrive at $$1=\frac{d(x)}{d(x^2)}\cdot 2x.$$ Finally $$\frac{d(x)}{d(x^2)}=\frac{1}{2x}$$ for $x\ne 0$.

szw1710
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  • Please have a look at https://math.stackexchange.com/questions/2294788/find-partial-derivative-of-a-1xb-1y-with-respect-to-a-2xb-2y – Gaurav Gupta May 24 '17 at 11:14
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Write $u = x^2$. If $x > 0$ then $x = \sqrt u$ and

$$\frac{\partial\ }{\partial x^2} x = \frac{\partial\ }{\partial u} \sqrt u = \frac{1}{2\sqrt u} = \frac{1}{2x}$$

If $x < 0$ then $x = -\sqrt u$ and the derivative remains the same.

Thus for all $x \neq 0$,

$$\frac{\partial\ }{\partial x^2} x = \frac{1}{2x}$$

The derivative at $x = 0$ does not exist.

Simon S
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  • Thanks for the answer. Please have a look at https://math.stackexchange.com/questions/2294788/find-partial-derivative-of-a-1xb-1y-with-respect-to-a-2xb-2y – Gaurav Gupta May 24 '17 at 11:15
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Hint :

Derivative of $f(x)$ w.r.t. $g(x)$ is :

$$\frac{\partial f(x)}{\partial \big(g(x)\big)}=\frac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}$$

Jaideep Khare
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  • Thanks for the answer. Please have a look at https://math.stackexchange.com/questions/2294788/find-partial-derivative-of-a-1xb-1y-with-respect-to-a-2xb-2y – Gaurav Gupta May 24 '17 at 11:15