Beginners ask questions about the derivation rule of higher numbers Isn't there a formula called the derivative of the inverse function = 1 / (the derivative of the original function). If the original function = exponential function (based on e), then the inverse function = natural logarithm If the derivative of the inverse function = 1 / X and the derivative of the original function = e ^ x, then 1 / x = e ^ (- x) is contradictory,

Beginners ask questions about the derivation rule of higher numbers Isn't there a formula called the derivative of the inverse function = 1 / (the derivative of the original function). If the original function = exponential function (based on e), then the inverse function = natural logarithm If the derivative of the inverse function = 1 / X and the derivative of the original function = e ^ x, then 1 / x = e ^ (- x) is contradictory,

No contradiction, you're missing a step substitution. For example, x = e ^ y, its inverse function is equal to iny and also equal to the reciprocal of the derivative of the original function, that is, 1 / e ^ y, and because x = e ^ y, it is equal to 1 / X

Simple derivative calculation and integral calculation Suppose the derivative of q3-10q2 + 25Q + 150 is 3q2-20q + 25. How does the derivative calculation work In addition, if you know how to integrate the quadratic equation after derivation to calculate the cubic equation before derivation, I know that the constant must be indefinite after calculation, and I can't understand the derivation formula

The n-th derivative of X is equal to N times the N-1 power of X
The derivative of the number is 0
Here, the derivative of the third power of Q is 3 times the second power of Q, and so on
Integration is the opposite of derivation
The n-th power integral of X is equal to N + 1 times the N + 1 power of X
The integral of a number is equal to the number times X

Derivation operation y=In(1+cosx)-[1/(1+cosx)]

In(1+cosx)-[1/(1+cosx)]'=[-sinx/(1+cosx)]-[sinx/(1+cosx) ²]
=(-2sinx-sinxcosx)/(1+cosx) ²

On the calculation of integral derivation Why is the derivative of the integral of F (T) on 0 to x f (x) instead of F (x) - f (0)?

If an original function of F (x) is f (x), then f (x) '= f (x),
So: (f (T) integral over 0 to x) '= (f (x) - f (0))' = (f (x)) '- (f (0))' = f (x)

Formula derivation calculation （x ²— 1） ³+ Derivative of 1 (find the cube of the difference between X square minus "1" plus the derivative of "1")

6X（x ²— 1) ²

E ^ (X-Y ^ 2) derivation How to find the derivative of e ^ (X-Y ^ 2),

E ^ (X-Y ^ 2) = 0, which is the derivation of the implicit equation: [e ^ (X-Y ^ 2)] '= 0 → e ^ (X-Y ^ 2) · (1-2y · y') = 0 ∵ e ^ (X-Y ^ 2) ＞ 0, ∵ 1-2y · y '= 0, then y' = 1 / (2Y). E ^ (X-Y ^ 2) = 0, then X-Y ^ 2 = 1. Then y = √ (x-1), then y '= 1 / (2 √ (x-1))