Thank you for y = (2 LNX + x ^ 2) / (3lnx + x ^ 2) derivation process?

Thank you for y = (2 LNX + x ^ 2) / (3lnx + x ^ 2) derivation process?

Using logarithm derivation: right
y = (2 ln x + x^2)/(3lnx + x^2)
Take logarithms on both sides and get the
lny = ln(2 ln x + x^2) - ln(3lnx + x^2),
We get the derivative of X on both sides
y'/y = (2/x+2x)/(2 ln x + x^2) - (3/x+2x)/(3lnx + x^2),
So the answer is
y' = y[(2/x+2x)/(2 ln x + x^2) - (3/x+2x)/(3lnx + x^2)],
It's up to you to simplify

If y = f (x + y) and f (x) is derivable and its derivative is not equal to 1, then dy / DX=

dy/dx=f'(x+y)+f'(x+y)dy/dx dy/dx=f'(x+y)/1-f'(x+y)

F (x) = xsin1 / x x x is not equal to 0
Shouldn't LIM (sin1 / x) / (1 / x) be equal to 1? Why is the derivative 0?

If t = 1 / x, then t tends to infinity, and the original is reduced to Sint / T, because t is infinite and Sint is between 0-1, so it is 0

Given that the square of (x + y) is 12 and the square of (X-Y) is 8, find the sum of the squares of X + y and XY

(x + y) & sup2; = 12 x & sup2; + Y & sup2; + 2XY = 12 1 formula (X-Y) & sup2; = 8 x & sup2; + Y & sup2; - 2XY = 8 2 Formula 1 formula + 2 Formula: X & sup2; + Y & sup2; = (12 + 8) △ 2 = 101 formula - 2 Formula: 4xy = 4 xy = 1 answer: X & sup2; + Y & sup2; = 10 xy = 1

Given x + 1 / x = 2, find the square of x minus 1 / 2 of X

The square of (x + x 1 / 2) is equal to 4, so that the square of x plus x 1 / 2 is equal to 2, and the square of x minus x 1 / 2 is equal to 0

Cosa / 2 = (2 √ 5) / 5, why cosa = 2cos quadratic A / 2-1 = 3 / 5

Cos a = cos (A / 2 + A / 2) = cos (A / 2) cos (A / 2) - sin (A / 2) sin (A / 2) = cos & # 178; (A / 2) - Sin & # 178; (A / 2) and COS & # 178; (A / 2) + Sin & # 178; (A / 2) = 1, so cos & # 178; (A / 2) - Sin & # 178; (A / 2) = 2cos & # 178; (A / 2) - 1, that is, cos a = 2cos & # 17

Simplification (1 + cosa + cos2a + cos3a) / (2cos ^ 2A + cosa-1)

(1+cosa+cos2a+cos3a)/(2cos^2a+cosa-1)=[(cosa+cos3a)+1+cos2a]/(2cos^2a+cosa-1)=[2cos(a+3a)/2*cos(a-3a)/21+cos2a]/(2cos^2a+cosa-1)=[2cos2a*cosa+1+cos2a]/(2cos^2a+cosa-1)=(2cos2a*cosa+2cos^2a]/(2cos^2a+c...

The fourth power of cosa minus 1 / 4, the square of cos2a minus 1 / 2, cos2a = how to do

∵ cos & # 178; a = 1 / 2 (1 + cos2a) (the power reduction formula of the variation of the double angle formula) ∵ cos & # 8308; a = [1 / 2 (1 + cos2a)] &# 178; ∵ cos & # 8308; A-1 / 4 * cos & # 178; 2a-1 / 2cos2a = 1 / 4 (1 + cos2a) &# 178; - 1 / 4 * cos & # 178; 2a-1 / 2cos2a = 1 / 4 (1 + 2cos2a + cos & # 178; 2a) - 1 /

What is the square a of (1-cos2a) sin2a / 2cos 2A sin?

Square a of (1-cos2a) sin2a / 2cos 2A sin
= (1-2cos^2 + 1) sin2a / (2cos2a sin^2a)
= 2sin^2a sin2a / (2cos2a sin^2a)
= tan2a

2cos ^ a = 1-cos2a, right
(1+sin40+cos40)/(1+sin40-cos40)
How much is the problem

Wrong, it should be
2cos^a=cos2a+1