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Since the integral variable is t, the integral is proposed by f(x) as a constant, so the integral 0 to 1 and 1dt is 1, so this equation is equal to f(x).
Derivative, also known as derivative.
Value. Also known as micro-business, it is calculus.
important foundational concepts in . When the function y=f(x) is an independent variable.
When x produces an increment δx at a point x0, the ratio of the delta δy of the output value of the function to the delta δx of the independent variable is at the limit a when δx approaches 0, if it exists, a is the derivative at x0 and is denoted as f'(x0) or df(x0) dx.
Common derivative formulas:
1. y=c (c is a constant) y'=0
2、y=x^n y'=nx^(n-1)
3、y=a^x y'=a^xlna,y=e^x y'=e^x4、y=logax y'=logae/x,y=lnx y'=1/x5、y=sinx y'=cosx
6、y=cosx y'=-sinx
7、y=tanx y'=1/cos^2x
8、y=cotx y'=-1/sin^2x9、y=arcsinx y'=1/√1-x^2
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y= ∫(x->a) f(t) dt
∫(a->x) f(t) dt
y' = -f(x)
Let f(x)=[ (0,x)xf(t)dt]f(x)=[ (0,x)xf(t)dt]=x* (0,x)f(t)dtf'(x) = (0,x)f(t)dt+x*f(x) Because it is a derivative of x, it is the independent variable of the function, not the integral variable of the integral, and it must be put outside, otherwise it is not easy to find. Of course, x is equivalent to a constant relative to the integral, and it can also be taken outside.
Derivatives are local properties of functions. The derivative of a function at a point describes the rate of change of the function around that point. If both the independent variables and the values of the function are real, the derivative of the function at a point is the tangent slope of the curve represented by the function at that point.
The essence of derivatives is to perform a local linear approximation of a function through the concept of limits. For example, in kinematics, the derivative of the displacement of an object with respect to time is the instantaneous velocity of the object.
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Let tx=s,xdt=ds
t=0,s=0
t=1,s=x
So. Original cavity = (0,x) f(s)1 x ds1 x (0,x) f(s) ds derivation. Derivative = -1 x square to silver (0,x) f(s) ds +f(x) x
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Parse: We know y'=dy/dx.
In other words, dy dx means to derive y!
Now d dx is followed by a definite integral, which means to find a derivative of a definite integral, the definite integral is a constant, and the derivative of the constant function is 0!
If d dx is followed by an indefinite integral, e.g. d dx f(x)dx, what is the result? We can do this by letting the original function of f(x) be f(x) c, then f(x) c f(x)dx, then d dx f(x)dx d dx f(x) c f'(x) 0 f(x), i.e. d dx f(x) dx f(x).
Note: Don't confuse definite integrals with variable upper limit integrals, definite integrals are constants, and variable epithelial explicit integrals are functions!
What you add is the variable upper bound integral: d dx (0,x)f(t)dt=f(x), and the derivative rule is, just replace the t in the integrand with the upper limit x. For example: d dx (0,x)sintdt=sinx
However, if the upper limit is not x, but some other function, such as x 2, then you have to multiply x 2 by the derivative of x 2 instead of t, i.e. by 2x, e.g., d dx (0,x 2)sintdt=sinx 2*2x=2xsinx 2
To give you a formula for macropei: (x),g(x)) f(x)dx f(g(x))*g'(x)-f(ψ(x))*x).
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f(x)= x,1)t 2lntdt.
First, use the partial integration method: Ye Zhengbi lntdt=tlnt- tdlnt=tlnt- t(1 t)dt=tlnt- dt=tlnt-t, and then substitute the upper and lower limits of the integral into the calculation of the rent clearance integral to obtain: f(x)=xlnx-x-(1ln1-1)=xlnx-x+1 to derive:
f'(x)=lnx+x(1/x)-1=lnx+1-1=lnx
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Yes, let f(t) = (t-1)f(t).
After the derivative of the frank plexus modulus is obtained by the integral let cherry blossom variable line function.
f(x) is (x-1)f(x).
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The derivative of the silver-lead integral of the variable limit beam = substituting the upper limit value by the integrand and multiplying by the upper derivative) - (the substituting the lower limit value by the lower limit value and multiplying the lower limit derivative).
The answer to this question, x*(x 2+1) f(x) is to directly replace the t in the integrand with x.
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The xf(x) answer and proof are shown below.
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Solution: i) When m 0, m+1 m 2, and if and only if m=1, take the equal sign to get 2sinx 2 sinx 1 sinx=1x=2k + 2 >>>More
Solution: According to the problem, the value range of the graph is in the upper part of the abscissa x-axis: that is, the value of the positive rotation curve x is: 2k
Date of birth:
Gregorian calendar) in 2015. >>>More