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It is known that the m power of a is equal to 2, the n power of a is equal to 5, and the value of the M + n power of ball a
M + n power of a = a ^ MxA ^ n = 2x5 = 10
Note: A ^ m represents the m-th power of a!
Discontinuities must not be differentiable, but why can the discontinuities in piecewise functions be defined to find the derivatives of discontinuities
Let me tell you that there are two cases of piecewise function at the piecewise point: 1. The function is continuous at the piecewise point; 2. The function is discontinuous at the piecewise point
For "the function is continuous at the piecewise point", there are two cases (1, the function is differentiable at the continuous point, 2, non differentiable)
There is only one case (1, non differentiable) for "the function at the piecewise point is discontinuous"
You said, "but why can the discontinuity in a piecewise function calculate the derivative of the discontinuity through the definition?" this definition only calculates a form. Its limit either does not exist or the left and right limits are not equal. If you go deeper, you will find that the differentiable function must not contain the first type of discontinuity
The derivative must be continuous, and the discontinuity must not be derivative. But why can the discontinuity in the piecewise function be defined to calculate the derivative of the discontinuity If it is discontinuous and non differentiable, the point between piecewise functions is the discontinuous point. Why does it have a derivative? Isn't this in contradiction with the fact that discontinuity must be non differentiable and differentiable must be continuous? I'm confused. I can't tell,
Piecewise functions only have different expressions in different intervals, but the function values may still be the same at the discontinuities, such as
y= 2x (x>=0)
5x (x
How to prove the derivative of Cotx with the definition of derivative? I can't prove it by definition!
lim(△x→0)[cot(x+△x)-cotx]/[(x+△x)-x]
=lim(△x→0)[cos(x+l△x)sinx-sin(x+△x)cosx]/[(sin(x+△x)sinx)*△x ]
=lim(△x→0) -[sin[(x+△x)-x]/△x]*[1/[sin(x+△x)sinx]
=lim(△x→0) -(sin△x/△x)*[1/sin(x+△x)sinx]
lim(△x→0)sin△x/△x=1
=-1/(sinx)^2
Can high school students learn calculus by themselves? Because mathematics and physics need to use these, mathematical functions are not good, want to learn this. Is it OK? By the way, what kind of calculus should I learn?
It's too difficult for you to learn calculus in senior one. What is learning calculus? Calculus is the general term of differential and integral. We should learn nature from the foundation
Calculate the derivative of the function at the specified point according to the definition of derivative: y = cosx, at x = π / 4
△y=f(π/4+u)-f(π/4)=cos(π/4+u)-cosπ/4=-2sin(π/4+u/2)sin(u/2)
△x=u
y’=f’(π/4)
=(u→0)lim[-2sin(π/4+u/2)sin(u/2)/u]
=(u→0)lim[-sin(π/4+u/2)sin(u/2)/(u/2)]
=-sinπ/4
=-√2/2
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