The fifth power of F (x) = 1 / 3-The derivative of 4x

The fifth power of F (x) = 1 / 3-The derivative of 4x

f'(x)=5/3*x^4-4

Y = (x-1) derivative to the fifth power

y =(X^2-1)^5
y`=5(x^2-1)^4*2x
=10x(x^2-1)^4

It is known that n is a positive integer and the power of XN is equal to the value of 2. Find the power of 3N of 3 x minus the power of 2n of 4 x 2

The n-th power of three times 24

Factorization factor xn + 1-6xn + 5xn-1

The original formula = x ^ (n-1) (x - 6x + 5) = x ^ (n-1) (X-5) (x-1)

It is known that the monomials of M and n have the same degree of the sixth power of M, the fourth power of N, the third power of 4m ^ xn cube and the (- 25) power of m square n ^ y + 1
The value of the square of the algebraic formula X - 2XY + y

Your statement is extremely unclear. Please check the title

The tangent equation of the curve y is equal to the square of x plus 2 / 1 at point (1,1)

y=2x/(x^2+1)
y'=[2(x^2+1)-2x*2x]/(x^2+1)^2
At point (1,1)
x=1,y=1,y'=0
So the tangent equation is y = 1

High school derivative problem: solve the linear equation that passes through a point P (4,2) on the curve y = √ X and is perpendicular to the tangent passing through this point
RT I get y = - 4x + 18, if not, I get the process, thanks

y=√x
y'=1/2√x
k=y'(4)=1/2*√4=1/4
∵ perpendicular to tangent
∴k1*k=-1
k1=-4
What is the linear equation
y-2=-4(x-4)=-4x+16
Y = - 4x + 18 your answer is right

If the derivative of y = X3 at x = 0 is 0, is the X axis its tangent?

Because according to the geometric meaning of derivative, the derivative of function y = x ^ 3 at x = 0 is 0, that is, the tangent slope of the image of function y = x ^ 3 at x = 0 is 0, which can be directly obtained from the definition of tangent
If you have any doubt about this problem, you must think that a curve should be on the same side of its tangent. This is a misunderstanding caused by the tangent of a circle in middle school. In fact, there is no such restriction on the definition of tangent. Please refer to the definition of curve tangent in advanced mathematics

If the tangent at point P on the curve y = x2-3x is parallel to the X axis, then the coordinate of P is ()
A. （-32，94）B. （32，-94）C. （-32，-94）D. （32，94）

Y ′ = 2x-3, let y ′ = 0. That is, 2x-3 = 0, then x = 32. Substituting into the curve equation y = x2-3x, then y = - 94

The coordinates of the tangent point parallel to the x-axis on the curve y = x3-3x are______ ．

Let the tangent point coordinate be (m, n) ∵ y = x3-3x, ∵ y ′ = 3x2-3, when 3x2-3 = 0, we get: x = ± 1, ∵ the tangent point coordinate is (- 1,2) or (1, - 2), so the answer is: (- 1,2) or (1, - 2)