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Let $y=x^2$ and $y=t^4$ $$\frac{\mathrm{d}y}{\mathrm{d}x}=2x$$ $$\frac{\mathrm{d}y}{\mathrm{d}t}=4t^3$$

Now, $\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{\mathrm{d}}{\mathrm{d}t}\left(2x\right)=\frac{\mathrm{d}}{\mathrm{d}t}\left(2t^2\right)=4t$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left(4t^3\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left(4x^{\frac{3}{2}}\right)=6t$

Why am I getting $\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)$ is not equal to $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)$? Isn't it a law of calculus?

imnotmurtaza
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Time Traveler
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1 Answers1

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$$\frac{d}{dt}\frac{dy}{dx}=\frac{d}{dt}\frac{dy}{dx}\frac{dx}{dx}= \frac{d}{dx} \left(\frac{dy}{dx}\right)\frac{dx}{dt}=\frac{d^2y}{dx^2}\frac{dx}{dt}~~~(1)$$ Next $$\frac{d}{dx} \left(\frac{dy}{dt}\right)=\frac{d}{dx} \left(\frac{dy}{dt} \frac{dx}{dx}\right)=\frac{d}{dx} \left(\frac{dy}{dx} \frac{dx}{dt}\right)$$ $$=\frac{d^2y}{dx^2}\frac{dx}{dt}+\frac{dy}{dx}\frac{d}{dx} \left(\frac{dx}{dt}\right)=\frac{d^2y}{dx^2}\frac{dx}{dt}+\frac{dy}{dx}\frac{d}{dx} \left(\frac{dx}{dt}\right)\frac{dt}{dt}$$ $$=\frac{d^2y}{dx^2}\frac{dx}{dt}+\frac{dy}{dx}\frac{d}{dt} \left(\frac{dx}{dt}\right)\frac{dt}{dx}=\frac{d^2y}{dx^2}\frac{dx}{dt}+\frac{dy}{dx}\frac{d^2x}{dt^2} \frac{dt}{dx} .~~~~(2)$$

One may check the correctnes of (1) and (2) and their inequality by taking $x=t^2, y=t^2, y^2=x^3$, etc.

Z Ahmed
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