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Y = 1 / x + √ (x) find the derivative
See the derivation formula: ① (U + V) '= u' + V '
②(x^n)'=nx^(n-1) (n∈Q)
(1/x)'=-1/x ²
√(x) =（x^0.5）=1/2*x^(-1/2)
y=1/x+√(x)
y'=-1/x ²+ 1/(2√x)
Known: F (x) = (x-1) (X-2) (x-3)... (X-100), find the value of F '(3) (i.e. find the derivative of F (3)) Please write down a more detailed process
F '(x) = (x-1)' * (X-2) (x-3)... (X-100) + (x-1) (X-2) '* (x-3)... (X-100) + (x-1) (X-2) (x-3)' *... (X-100) + + (x-1) (X-2) (x-3)... (X-100) 'except for (x-1) (X-2) (x-3)' *... (X-100), there is an x-3 term, so x = 3 is equal to 0 and (
What kind of function is a composite function Does y = (2x + 3) ^ 2 count? Does r = (3V / 4 π) ^ (1 / 3) count? Does y = (x + 1) ^ 99 count? If not, why? There is also the question of derivation 1) Y = (2x + 3) ^ 2 using the formula F (g (x)) = f (x) '* g (x)' of the composite function, the result is y '= 8x + 12 Then suppose 2x + 3 is regarded as a whole, and the result obtained according to the derivation formula of the power function is y '= 4x + 6 2) Y = (x + 1) ^ 99 the result obtained by the formula of the composite function is y '= 99 (x + 1) ^ 98 assuming that x + 1 is regarded as a whole, the result obtained by the derivation formula of the power function is also y' = 99 (x + 1) ^ 98. Why do the two different situations 1) and 2) occur? Is it because there is a coefficient in front of X in 1)? There is also a question about the derivation of the function r = (3V / 4 π) ^ (1 / 3). If it is simply regarded as a power function, the result of derivation should be r '= (1 / 3) (3V / 4 π) ^ (- 2 / 3), which is also the correct answer But I regard it as a composite function. Let u = 3V / 4 π, and then use the derivative of the composite function to find the result, which becomes R '= (1 / 4 π) (3V / 4 π) ^ (- 2 / 3). What's wrong?
The so - called compound function is that you artificially select an intermediate variable
For example, y = (2x + 3) ²
I choose an intermediate variable U = 2x + 3
Then the primitive is equivalent to
y=u ²= f(u)
u=2x+3=g(x)
(the above two expressions are enclosed in curly brackets)
The requirements are:
y′=f′(u)·g′(x)
= 2u·2
= 8x+12
In your first question, you forgot to take the derivative of 2x + 3 = u over X,
That is, G '(x) is omitted, and obviously G' (x) = u '= 2
And you did the second question right just because G '(x) = (x + 1) ′ = 1
It's the same whether you take it or not
Question 3... What's wrong? It's all wrong
In fact, derivation is not difficult. You can do it right as long as you step by step
On the derivative of marginal utility. However, I still hope experts can explain it For example, u = XY. MUX = y But if u = x + y, How do we get MUX?
Think of Y as a constant, so μ X=1
Mathematical derivation formula Y=M/N Find the derivative of this equation
y=m/n
y'=(m'*n-m*n')/(n^2)
Derivative this formula What happens after the derivation of the formula y = 3T ^ 3 + X
If you take the derivative of X and t is a constant, then
y'=1
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