Let y = 3x + 1, then the derivative X '(y) of its inverse function x = x (y)=

x=(y-1)/3=y/3-1/3
x'(y)=1/3

Find the derivative of inverse function of y = sinhx In addition, what are the specific steps of the derivation of the inverse function? Do you want to find the inverse function of the function first? If the inverse function has been found, it is not good to directly use the inverse function to obtain the derivative. What is the significance of that formula? According to the formula, the derivative of the inverse function is the reciprocal of the derivative of the original function. Isn't the answer to this question 1 / coshx? But that's not the answer

You have misunderstood the derivation theorem of inverse function. It is not the reciprocal of the derivative of the inverse function, but the reciprocal of the derivative of the direct function corresponding to the inverse function
This theorem can be used or not. In fact, it is not contradictory. The convenience of using this theorem is that you can find the derivatives of some functions that are difficult to derive. There are also the derivatives of functions that are not easy to be obtained by inverse functions. For example, y = arcsinx, the derivative of this function is similar to that of anti trigonometric function. You can't directly and easily calculate the derivative of this function
It is easy to find the inverse function and the derivative of the inverse function
Therefore, we should divide the situation into different parts. We should not adopt a one size fits all approach

How to solve the derivative of inverse function of y = x + LNX

y'=1+1/x
The derivative of the inverse function is: y = 1 / y '= 1 / (1 + 1 / x) = x / (x + 1)

LNX and the x power of E are inverse functions of each other. Why is the derivative of LNX not equal to the reciprocal of the derivative of X of E?

In fact, it can be seen that y = f (x) and x = f (y) are both derivations of X: y '= f' (x) 1 = f '(y) * y' (the rule for the derivation of composite functions). Here we can see that two Y 'are reciprocal to each other, but you have to look at the two f's

Why is an inverse function and its original function reciprocal in derivative?

Because they're down against each other

It is said that the derivative of an inverse function is the reciprocal of the derivative of a direct function, but in this example: y = X2, its inverse function is x = √ y, and its derivative is y = 1

You are very talented. I remember this is the conclusion of the textbook,
Your two functions are not reciprocal functions. The inverse function requires them to define and range exactly opposite
Only when they are inverse functions can we use this conclusion
Engineers and scientists should pay attention to the applicable conditions of conclusions
The derivation of inverse function is monotone, continuous and differentiable

The derivative of a function is reciprocal to that of its inverse function So why does the derivative of e ^ X and the derivative of ㏑ x not work

No, you are wrong. It should be that the derivative of e ^ X and the derivative of LNY (y = e ^ x) (or the derivative of e ^ y (y = LNX) and the derivative of LNX) are reciprocal
(LNY) '= 1 / y = 1 / e ^ x (or (e ^ y)' = e ^ y = e ^ (LNX) = x)

The derivative of the inverse function is equal to the reciprocal of the derivative of the direct function, such as y = e ^ X and y = LN

y＇﹙x﹚＝e^x＝1/x＇﹙y﹚＝1/﹙1/y﹚

Derivative of inverse function y = (1 / 2) ln ((1-x) / (1 + x))

The inverse function of y = (1 / 2) ln ((1-x) / (1 + x)) is hyperbolic tangent y = th (x), and its derivative is y = 1 / CH (x) ^ 2
Please refer to higher mathematics of Tongji University for details
Hyperbolic cosine CH (x) = (exp (x) + exp (- x)) / 2
Hyperbolic tangent th (x) = (exp (x) - exp (- x)) / (exp (x) + exp (- x))

Find the derivative of the inverse function of y = x-ln (1 + x)!

y=x-ln(1+x)
y'=1-1/(1+x)*1
=(1+x-1)/(1+x)
=x/(1+x)
Derivative of inverse function: y '= (1 + x) / X
=1/x+1

Let y = 3x + 1, then the derivative X '(y) of its inverse function x = x (y)=