/3x + 4 / and / y + 2 / are opposite to each other. How much is x + y I did the same thing, but the answer is 9

/3x + 4 / and / y + 2 / are opposite to each other. How much is x + y I did the same thing, but the answer is 9

Both absolute values of this question must be 0
So x = - 4 / 3, y = - 2
So x + y = - 10 / 3

Finding the derivative of y = ln (sin2x + 2 ^ x)

y=ln(sin2x+2^x)
y'=1/(sin2x+2^x) * (sin2x + 2^x)'
=1/(sin2x+2^x) * (cos2x * (2x)' + ln2 * 2^x)
=(2cos2x + ln2 * 2^x) / (sin2x+2^x)

Derivatives of y = (ln2x) ^ 3 and D (sin2x)

y=(ln2x)^3
y'=3(ln2x)²*(ln2x)'
=3(ln2x)²*1/(2x)*(2x)'
=3(ln2x)²*1/(2x)*2
=3(ln2x)²/x
d(sin2x)
=cos2xd(2x)
=2cos2xdx

Derivative of y = (sin2x) ^ 2

y'=2sin2x(sin2x)'=2sin2xcos2x(2x)'=4sin2xcos2x=2sin4x

The 50th derivative of y = (x ^ 2 + 2x + 1) sin2x
Find the value of the 50th derivative of y that brings π in

The higher derivative of multiplication of two functions: use the formula. Y Λ (n) = {C (upper I, lower n) * f (x) Λ (I) * g (x) Λ (n-i) [find the sum of I = 0 to I = n]}. Find the answer

If the derivative f (x) = 2x (x + 1) ^ 2, then the extreme value of function f (x)

What he gave is the derivative formula. It's just that the apostrophe on f (x) can't be found, and the square is only on (x + 1), and there's no square for 2x

The function f (x) is continuous. When x tends to 0, the limit of F (x) / X is 2, y = the derivative of F (x) at x = 0

The limit of F (x) / X is 2
because
Limx=0
therefore
lim(x->0)f(x)=0
And the function is continuous, so
lim(x->0)f(x)=f(0)=0
therefore
lim(x->0)f(x)/x=lim(x->0)[f(x)-f(0)]/(x-0)=f'(0)=2
Namely
The derivative of y = f (x) at x = 0 f '(0) = 2

Let f (x) be differentiable in a finite interval (a, b), and the right limit of F (x) at a point is infinite. Can we judge that the right limit of the derivative of F (x) at a point is also infinite?

f(x) = lnx + sin(lnx),a=0

I can't calculate the derivative of (Sin & # X / 2)

sin²(x//2)=(1-cosx)/2
So (Sin & # 178; X / 2) "= (SiNx) / 2

The derivative X of F (x) = x ^ 2 * sin (1 / x) is greater than 0. Why do I use definition method and formula method to calculate it differently?
Please help me to work out the detailed steps

When you use the definition to calculate - cos (1 / x) x - > 0, 1 / X - > infinite COS is constantly fluctuating, right? It's not a stable value, so the result is wrong. You need to use the definition to calculate 0