This may be a silly question but for example if you had the gradient at $x=4$ of $y=x^2+1$, then can you just calculate $\frac{dx}{dy}$ by finding $\frac{dy}{dx}$ and flipping it over? Or must you make $x$ the subject and differentiate?
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1That's not a silly question at all. It's a completely natural question. – layman Mar 13 '17 at 10:15
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Related: Confusion about integrals. – Andrew D. Hwang Mar 13 '17 at 10:57
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Be careful when you move on to multivariable calculus, though: you generally can’t do the same thing with partial derivatives. – amd Mar 13 '17 at 18:41
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If $f$ is invertible and differentiable, with $f'$ never zero, then $f^{-1}$ is also differentiable and
$$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}. $$
Does this help?
Stefano
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implicit differentiation operator d/dy
$y = x^2 + 1$
$d/dy(y) = d/dy(x^2 + 1)$
$1 = 2x dx/dy$ (by the chain rule on the RHS)
$dy / dx = 2x$
as expected
Cato
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Yes, you can. If both derivatives exist, they are reciprocal, that is
$\frac{dy}{dx}\frac{dx}{dy}=1$
learner
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