Let f (x) = x (X-2) (x-3), find f '(x). Find the tangent equation and normal equation of the curve with y = x ^ 5 at point (1,1) Find the maximum and minimum value of function f (x) 2x ^ 3-3x ^ 2 in closed interval [- 1,4] High numbers can't find answers

Let f (x) = x (X-2) (x-3), find f '(x). Find the tangent equation and normal equation of the curve with y = x ^ 5 at point (1,1) Find the maximum and minimum value of function f (x) 2x ^ 3-3x ^ 2 in closed interval [- 1,4] High numbers can't find answers

f(x)=x(x-2)(x-3)=x^3-5x^2+6x
f'(x)=3x^2-10x+6
y=x^5
Y '= 5x ^ 4, when x = 1, y' = 5, that is, the slope is 5, so the tangent equation is y = 5x-4, and the normal equation is y = - 1 / 5 x + 6 / 5
f(x)=2x^3-3x^2
F '(x) = 6x ^ 2-6x = 6x (x-1), so x = 0 and x = 1 are the two extreme points of F (x), f (0) = 0, f (1) = - 1, f (- 1) = - 5, f (4) = 80, so the maximum value of F (x) is 80 and the minimum value is - 5 in the closed interval [- 1,4]

The general solution of the differential equation y ″ + y = - 2x is______ ．

The characteristic equation corresponding to the homogeneous equation y ″ + y = 0 is: λ 2 + 1 = 0, then the characteristic root is: λ 1, 2 = ± I, and its general solution is. Y = c1cosx + c2sinx, because the non-homogeneous term is: F (x) = - 2x = - 2xe0, and λ = 0 is not the characteristic root, so we can set the special solution of the non-homogeneous equation as: y * = a + BX, substituting into the original equation, we can get

The reciprocal of higher number y = y + XY solves the differential equation

Chapter 12 of advanced mathematics

Find the differential of the following function: xy = a ^ 2

xy=a^2
ydx+xdy=0
dy=-y/x dx
dy=-a²/x² dx

Finding the differential of the function determined by xy = e ^ XY

Consider y as a function of X, and derive x on both sides of the equal sign
y + xy' = (y + xy')e^xy
It is proposed that y ': X (1 - e ^ XY) y' = y (e ^ XY - 1)
Because X and y are not 0 (the exponential function at the right end of the equal sign is always greater than 0, so the left end is not 0), so (1 - e ^ XY) is not 0
y' = y/x

Derivation of y = ln TaNx / 2

This is the derivation of a composite function. Remember a formula [V (U)] ' = u & #39; V & #39; (U) that is ln (TaNx / 2) + ln1 / 2 = (1 / (2cosx ^ 2)) * (2 / TaNx), where ln1 / 2 is a constant and the derivative is zero

Y = ln (secx + TaNx). I can't figure it out
Method 1: y '= [1 / (secx + TaNx)] * (secxtanx + sec & # 178; x) = (secxtanx + sec & # 178; x) / (secx + TaNx) = secx (secx + TaNx) / (secx + TaNx) = secx
But method 2: secx + TaNx = TaNx / 2, then y '= lntanx / 2 = [1 / (TaNx / 2)] * sec & # 178; X * 1 / 2 = CSCX
What's the situation?

Where does secx + TaNx = Tan (x / 2) come from? It's obviously wrong
Substituting x = 0 into 1 = 0
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Quadratic derivation of Ln (TaNx)

Let f (x) = ln (TaNx)
be
f'(x)=1/tanx *(tanx)'
=1/tanx *1/cos^2 x
=cosx/sinx *1/cos^2 x
=1/(sinxcosx)
=2/sin2x
f''(x) =-2 *2*cos2x/sin^2 2x
=-4cos2x/sin^2 2x

Find the derivative of the following function: (1) y = e ^ x (SiNx -- cosx) (2) y = (1 + sin2x) ^ 4

（1）y'=e^x(sinx-cosx)+e^x(cosx+sinx)=2e^xsinx
（2）y'=4(1+sin2x)³(1+sin2x)'
=8cos2x(1+sin2x)³
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Finding the differential of e to the power of x-sin2x

dy=(e^x-2cos2x)dx