Derivative of e to the KX power

Derivative of e to the KX power

Because the derivative of e ^ x is still e ^ X. if x here becomes KX, we have to use the chain law and multiply it by the derivative of KX, which is K
∴d/dx(e^kx)=(e^kx)*k

F (x) = x * e ^ - x belongs to R

f(x)=x*e^(-x)
First find [e ^ (- x)] '= [e ^ (- x)] * (- x)' = - e ^ (- x)
Therefore, f '(x) = x' * e ^ (- x) + X * [e ^ (- x)] '= e ^ (- x) + X * (- e ^ (- x)) = e ^ (- x) * (1-x)

F (x) = e ^ X / (x-1) derivative of X

General [f (x) / g (x)] '= [f' (x) g (x) - f (x) g '(x)] / [G ^ 2 (x)]
Therefore, for this topic f '(x) = [e ^ x * (x-1) - e ^ x * 1] / (x-1) ^ 2 = e ^ x * (X-2) / (x-1) ^ 2

Such as the title! If the function g (x) = f (x) + 2 / X is a subtractive function on [1,4], I'm anxious to find the value range of real number a, Known function f (x) = x + alnx When a = 2E, find the monotone interval and extreme value of function f (x) If the function g (x) = f (x) + 2 / X is a subtractive function on [1,4], find the value range of real number a I was too anxious just now

First question: take the derivative of F (x) and get f '(x) = 2x + 2E / X
This is a cross check function, which is greater than 0 in [1,4], so f (x) is added
So f (1) = 1 is the minimum and f (4) = 16 + 2 * e * ln4 is the maximum
Second question: take the derivative of G (x) and get G '(x) = 2x + A / X-2 / (x ^ 2)
Make g '(x)

It is known that the odd function f (x) is a subtractive function on the definition field [- 2,2]. If f (2a + 1) + F (4a-3) > 0, find the value range of the real number a

Because f (x) is an odd function,
Therefore, f (2a + 1) + F (4a-3) > 0 can be transformed into f (2a + 1) > - f (4a-3) = f (3-4A),
And f (x) is the subtraction function on the definition field [- 2,2],
So there are
2a+1<3−4a
−2≤2a+1≤2
−2≤4a−3≤2 , Solution 1
4≤a<1
3,
So the value range of real number a is 1
4≤a<1
3.

It is known that the odd function f (x) is a subtractive function on the definition field [- 2,2]. If f (2a + 1) + F (4a-3) > 0, find the value range of the real number a

Because f (x) is an odd function,
Therefore, f (2a + 1) + F (4a-3) > 0 can be transformed into f (2a + 1) > - f (4a-3) = f (3-4A),
And f (x) is the subtraction function on the definition field [- 2,2],
So there are
2a+1<3−4a
−2≤2a+1≤2
−2≤4a−3≤2 , Solution 1
4≤a<1
3,
So the value range of real number a is 1
4≤a<1
3.