How to derive: SiNx ^ cosx; X ^ (1 / x) How to derive: Question 1: SiNx ^ cosx; Question 2: x ^ (1 / x) I don't seem to have a formula with X in the index~

How to derive: SiNx ^ cosx; X ^ (1 / x) How to derive: Question 1: SiNx ^ cosx; Question 2: x ^ (1 / x) I don't seem to have a formula with X in the index~

【1】 Let y = SiNx ^ cosx take logarithms on both sides, then LNY = cosx * ln (SiNx) + cosx * 1 / SiNx * cosxy '= y * [- SiNx * ln (SiNx) + cosx * 1 / SiNx * cosx] y' = SiNx ^ cosx * [- SiNx * ln (SiNx) + (cosx) ^ 2 / SiNx] [2] let y = x ^ (1 / x) be the same

Derivation cos (cosx)

solution
Composite function
y=cosu, u=cosx
y'=-sinu, u'=-sinx
So y '= [sin (cosx)] SiNx

(y + x = LNA))

[1 / (a + X under root sign) + 1] / (a + X under root sign)

Derivative y = ln (- x)

A:
y=ln(-x)
y'(x)=[1/(-x)]*(-1)=1/x
So:
The derivative of y = ln (- x) is y '(x) = 1 / X

The lower root y = LNY = 1 LNX = 1 LNX: 1

1,y=ln(1-x)y'=1/(1-x)*(1-x)'=1/(1-x)*(-1)=1/(x-1);2,y=ln [1/√(1-x)]=-ln √(1-x)y'=-1/√(1-x)*[√(1-x)]'=-1/√(1-x)*[(1/2)*1/√(1-x)]*(1-x)'=-1/√(1-x)*[(1/2)*1/√(1-x)]*(-1)=1/[2(1-x)]3,y=ln √(1-x)y...

What is the derivative of derivative ~, y = 2 ^ x * ln x? 3Q

y=2^（xln x） y'=ln2(lnx+1)*2^（xln x）

LNX derivative is 1 / x, so what is the derivative of Ln (x + y)? Why is the derivative of X over u DX / dt = 2T / (1 + T ^ 2) guys... Can you give me a detailed answer....

If we want to find the derivative of Ln (x + y), first of all, the title must be clear, what is y, where y is a function of X, is it just a constant, or is it a variable independent of X
By convention, y should be a function of X, so the derivative of Ln (x + y) is (y '+ 1) / (x + y)
dx/dt=(1+t^2)'/(1+t^2)=2t/(1+t^2)

Derivative: y = ln [(x ^ 4) / √ (x ^ 2 + 1)]

y=ln[(x^4)/√(x^2+1)]
∴y'={1/[(x^4)/√(x^2+1)]}*[4x^3√(x^2+1)-x^4*(1/2)2x/√(x^2+1)]/[√(x^2+1)]^2
=[√(x^2+1)/x^4]*[(3x^5+4x^3)/√(x^2+1)]/(x^2+1)
=(3x^5+4x^3)/[x^4(x^2+1)]
=(3x^2+4)/(x^3+x)

Ln (x ^ 2 + y ^ 2) = arctan (Y / x) derivation Implicit Differentiation

Derivation of X on both sides of the equation
(2x+2yy')/(x^2+y^2)=[(xy'-y)/x^2]/[1+(y/x)^2]
2(x+yy')/(x^2+y^2)=(xy'-y)/(x^2+y^2)
2x+2yy'=xy'-y
(x-2y)y'=2x+y
y'=(2x+y)/(x-2y)

Derivation y = ln cos (2x + 1)

y=ln cos(2x+1)
y'=[ln cos(2x+1)]'
=1/cos(2x+1)*[cos(2x+1)]'
=1/cos(2x+1)*[-sin(2x+1)]*(2x+1)'
=-2/cos(2x+1)*sin(2x+1)
=-2tan(2x+1)