Finding the derivative of y = f (e ^ - x) Such as the title

Finding the derivative of y = f (e ^ - x) Such as the title

Y'=-f'e^(-x)
-e^-x*f(x)'
The answer to tassel is positive
Note: where f is f (e ^ - x)
Given that f (x) is an odd function defined on R, and when x > 0, f (x) = X3 + 2x-3, the analytic expression of F (x) on R is obtained
x> 0, f (x) = x ^ 3 + 2x-3
F (x) is an odd function defined on R
According to the definition of odd function: F (- x) = - f (x)
∴f（-x)=-(-x^3-2x-3)=x^3+2x+3
That is, when ＜ 0, f (x) = x ^ 3 + 2x + 3
The analytic expression of F (x) on R is as follows:
f(x)=x3+2x+3 x＜0
f(x)=x3+2x-3 x＞0
Let x0, (- x) ^ 3 + 2 (- x) - 3 = - x ^ 3-2x-3
X>0
-x < 0
f(x) = x^3+2x-3
f(-x) = - f(x) ( f is odd)
= - (x^3+2x-3)
= (-x)^3 +2(-x) + 3
y = -x 0
= x^3+2x+3 , for x< 0
when x =0 , f(x) = ???? ( I don't know)
Odd function f (0) = 0
X0
So f (x) is suitable for f (x) = x & sup3; + 2x-3
Then f (- x) = - X & sup3; - 2x-3
Odd function f (x) = - f (- x) = x & sup3; + 2x + 3
So f (x)=
x³+2x+3，x0
What are the derivatives of F (x) = ex-m-x? PS and x-m are the degree of E
F '(x) = ex-m-1, x-m is the degree of E
Given that f (x) is an odd function defined on R, and when x > 0, f (x) = X3 + X + 1, find the analytic expression of F (x)
X = 0 f (- 0) = f (0) = - f (0) f (0) = 0 x0 f (- x) = - x ^ 3-x + 1 = - f (x) f (x) = x ^ 3 + X-1 piecewise function f (x) = {F1 = X3 + X + 1, x > 0, F2 = 0, x = 0, F3 = x ^ 3 + X-1, X
f（x）＝x3 +x+1
What is the derivative of F (x) = ex / x
f(x)=[e^x]/(x)
f'(x)=[(e^x)'×(x)－(e^x)×(x)']/(x²)
=[xe^x－e^x]/(x²)
f(x)=[e^x]/(x)
f'(x)=[(e^x)'×(x)－(e^x)×(x)']/(x²)
=[xe^x－e^x]/(x²)
Given that f {1 / X} = 1 / x + 1, then the analytic expression and the domain of definition of function FX
Let's say f (1 / x) = 1 / (x + 1),
Let 1 / x = t, then x = 1 / T (t ≠ 0), f (T) = 1 / [(1 / T) + 1] = t / (1 + T),
Therefore, the function is f (x) = x / (1 + x), (x ≠ 0)
Let t = 1 / x, then t ≠ 0, then f (T) = t + 1, t ≠ 0,
The analytic formula F (x) = x + 1 is used to define all real numbers whose field is x ≠ 0.
What is the derivative of a function Sin & # x
∫sin²xdx
=∫(1-cos2x)/2 dx
=1/4∫(1-cos2x)d(2x)
=1/4(2x-sin2x)+C
It is known that f (x) is an odd function defined on R. when x is less than 0, f (x) = X3 + X-1, find the analytic expression of F (x)
∵ f (x) is an odd function
=X³+X-1（X＜0）
∴ F =-X³-X-1（X＞0）
=-1 (X=0）
F =-X³-X-1（X＞0）
=-1 (X=0）
Finding the derivative of y = TaNx & # 178
This is a composite function. First, take the square of X as the whole T, find the reciprocal of Tan T, that is, sin t divided by the reciprocal of cos t, and the result is cos t & # 178; one third, multiply by the reciprocal of T, that is, 2x. The answer is y '= 2x / cos X4
It is known that the odd function f (x) defined on R is a decreasing function. If X1 + x2 < 0, X2 + X3 < 0, X3 + X1 < 0, then the value of F (x1) + F (x2) + F (x3)______ .
∵ X1 + x2 ＜ 0, X2 + x3 ＜ 0, X3 + x1 ＜ 0, ∵ x1 ＜ - X2, X2 ＜ - X3, x3 ＜ - x1, and f (x) is a monotonically decreasing odd function defined on R, ∵ f (x1) > F (- x2) = - F (x2), f (x2) > F (- x3) = - f (x3), f (x3) > F (- x1) = - f (x1), ∵ f (x1) + F (x2) ＞ 0, f (x3) + F (x1) ＞ 0, ∵ the three formulas are summed up and sorted out（ X1) + F (x2) + F (x3) > 0, so the answer is: greater than 0