Is the derivative equal to one over the derivative of the inverse?

[Crosson]

Is it true that (dy/dx) = 1/(dx/dy) ? Is it still true if these are partial derivatives of a function of multiple variables? (All this, assuming that the inverse function exists)

Can anybody prove it in terms of the definition of a derivative? or any sort elementary proof at all? (I can almost convince my self it is true with Linear Algebra).

I haven't seen it proven in any books, but it is true for all of the functions I know of. Can I get a definitive answer?

 

[matt grime]

yes, dy/dx= 1/(dx/dy), when both are defined.

It doesn't make sense to ask about it for partial derivatives in the manner you do.

Suppose that y=f(x) implicitly defines y as a function of x. differentiate both sides wrt y

1 = df(x)/dy

apply the chain rule

1= (dx/dy) (df/dx)

or

1= (dx/dy)(dy/dx)

For partials you'd need to be more specific, are x and y both variables of this functions, or is, y=f(x,z,w...) the thing you're after? You can apply what I did above in the partial case for the second type.

 

[M98Ranger]

The statement that the derivative of a function is equal to one over the derivative of its inverse is not always true. This only holds for functions that are invertible and have a well-defined inverse function.

For a one-variable function, the statement can be written as: (dy/dx) = 1/(dx/dy). This is only true for functions that are one-to-one, meaning that each input value corresponds to a unique output value. In this case, the inverse function exists and the statement holds true.

However, for functions that are not one-to-one, the inverse function may not exist or may not be well-defined. In this case, the statement is not true.

This also applies to partial derivatives of functions with multiple variables. The statement (dy/dx) = 1/(dx/dy) is only true if the function is invertible and has a well-defined inverse function. Otherwise, it does not hold.

To prove this statement, we can use the definition of a derivative. The derivative of a function f(x) at a point x=a is defined as the limit of the difference quotient as h approaches 0:

f'(a) = lim(h->0) (f(a+h) - f(a))/h

Similarly, the derivative of the inverse function g(x) at a point x=b is defined as:

g'(b) = lim(h->0) (g(b+h) - g(b))/h

Now, if we let h = 1/f'(a), we can rewrite the first equation as:

f'(a) = lim(h->0) (f(a+1/f'(a)) - f(a))/1/f'(a)

Using the definition of the inverse function, we can replace f(a+1/f'(a)) with g(f(a)):

f'(a) = lim(h->0) (g(f(a)) - f(a))/1/f'(a)

Now, we can substitute this into the second equation:

g'(b) = lim(h->0) (g(b+h) - g(b))/h = lim(h->0) (f(a+1/f'(a)) - f(a))/1/f'(a) = f'(a)

Therefore, we can conclude that g'(b) = f'(a) = 1/(f'(a)), which proves the statement for invertible functions.

In conclusion, the statement (dy/dx) = 1/(dx/d