Given that the slope of the tangent at a point on the function f (x) = x ^ 2 is equal to 1, find its tangent equation

Given that the slope of the tangent at a point on the function f (x) = x ^ 2 is equal to 1, find its tangent equation

Analysis: ∵ the slope of the tangent at a point on the function f (x) = x ^ 2 is equal to 1
∴f‘(x)=2x=1==>x=1/2
f(1/2)=1/4
‡ a point (1 / 2,1 / 4)
The tangent equation passing through this point is Y-1 / 4 = 1 * (x-1 / 2) = = > y = X-1 / 4

Given that the tangent slope of the function f (x) = x ^ 2 + (1 / 2lnx-a) x + 2 at point (1, f (1)) is 1 / 2, find the value of A

F (x) = x ^ 2 + (1 / 2lnx-a) x + 2 derivative function f '(x) = [x ^ 2 + (1 / 2lnx-a) x + 2] ′ = 2x + 1 / 2lnx + 1 / 2-af (x) = x ^ 2 + (1 / 2lnx-a) x + 2 tangent slope at point (1, f (1)) is 1 / 2, that is, when x = 1, f' (1) = 2 + 1 / 2ln1 + 1 / 2-A = 5 / 2-A = 1 / 2, that is, a = 2

How to find the derivative of F (x) = x Λ 3?

f. (x) = square of 3x

F (x) = (x-1) (X-2) (x-3). (x-99) (X-100) find the derivative of F (1)

f′(x)=(x-1)′［(x-2)(x-3)(x-4)……(x-100)］+(x-1)［(x-2)(x-3)(x-4)……(x-100)］′ f(x=1)=1 ×［ (-1)(-2)(-3)……(-99)］+(1-1)［(x-2)(x-3)(x-4)……(x-100)］′ =-99!

Let f (x) = x (x-1) (X-2)... (X-100), then what is the derivative of F (0)

F = x * [(x-1)... (X-100)] find the derivative f '= [(x-1)... (X-100)] + X * @ do not need to find the following formula, substitute 0, f' (0) = 1 * 2 * 3.. + 0 * @ = 100! The answer is the factorial of 100

Find the 99th derivative of function y = x ^ 99 / (1-x). Fast

y=x^99/(1-x)=（x^99-1）/(1-x)+1/(1-x)
99th derivative of y = 99th derivative of 1 / (1-x)
z=(1-x)^(-1)
z'=(1-x)^(-2)
z''=2(1-x)^(-3)
99th derivative of y = 99! (1-x)^(-100)