The derivative of F (x) = sin (2 ^ X-2 ^ x-1)

The derivative of F (x) = sin (2 ^ X-2 ^ x-1)

f（x）=sin[2^x-2^(1/x)]f'(x)=cos[2^x-2^(1/x)]* [2^x-2^(1/x)]' =[2^x*ln2-2^(1/x)*ln2*(1/x)']*cos[2^x-2^(1/x)] =[2^x+1/x²*2^(1/x)]ln2*cos[2^x-2^(1/x)]

Finding the derivative of F (x) = sin (x / 2)

Let y = x / 2, then the original problem becomes to find the derivative of F (x) = siny, where y is a composite function, so f (x) = y '(siny)' = 1 / 2 * cos (x / 2)

Let f (x) = SiN x, then the derivative of F (f (x)) is?

cos（sin（x））cos（x）

F (x) = {(1 / x) * SiN x / 5, = 0, a, x = 0} find the derivative of x = 0?
When x is not equal to 0, y = (1 / x) * SiN x / 5
When x = 0, y = a

(1) X -- > 0, limf (x) = 1 / 5 (2) if a ≠ 1 / 5, then the function is discontinuous = = > the function is not differentiable at x = 0 (3) a = 1 / 5, X -- > 0, Lim [f (x) - f (0)] / x = Lim [sin (x / 5) / X-1 / 5] / x = Lim [sin (x / 5) - X / 5] / x ^ 2 = Lim [cos (x / 5) / 5-1 / 5] / (2x) Application of the law of rolpida = limsin (x

The problem of integral derivation: ∫ (upper limit x lower limit a) t * f (T) DT

xf(x)
If a is a constant, you don't need to subtract it
Unless the upper and lower limits are functions, not variables

Please explain the derivation process of integral derivation of [∫ (0, x) TF (T) DT] '= XF (x) -∫ (0, x) f (T) DT,

∵[∫(0,x) f(t)]' = f(x)
[∫(0,x) xf(t) dt]' = [x∫(0,x) f(t) dt]'
= x * [∫(0,x) f(t) dt]' + (x)' * ∫(0,x) f(t) dt
= x * f(x) + 1 * ∫(0,x) f(t) dt
= xf(x) + ∫(0,x) f(t) dt

The whole problem is: ∫ TF (T) DT = XF (x) + x ^ 2, (upper limit of integral is x, lower limit is t), find f (x)

The derivation is XF (x)
There is XF (x) = [XF (x)] '+ 2x
Both sides are multiplied by e ^ (- x) at the same time

Why is it necessary to put x outside the integral sign to seek the derivative of [∫ (0, x) XF (T) DT]?
0 is the lower limit and X is the upper limit

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 the derivative of X, which is the independent variable of the function, not the integral variable of the integral, and it must be put out, otherwise it is not easy to find

Derivation of integral TF (x-t) DT
Using the substitution method, x-t = u, t = x-u, but why is DT Du, not (- Du)

t=x-u
dt=d(x-u)=-du
you 're right
It should be DT = - Du

∫X^4/1+x dx.

∫x^4/(1+x)]dx
=∫[(x^4-1)+1]/(1+x)]dx
=∫(x^4-1)/(1+x)+∫1/(1+x)dx
=∫(x²+1)(x²-1)/(1+x)dx+∫1/(1+x)dx
=∫(x²+1)(x-1)dx+∫1/(1+x)dx
=∫(x^3-x²+x)dx-∫dx+∫1/(1+x)dx
=x^4/4-x³/3+x^2/2-x+ln(1+x)+C