The derivative of Ln (x ^ a) Such as the title Ln (x ^ 2 + 2) ^ (1 / 5)

The derivative of Ln (x ^ a) Such as the title Ln (x ^ 2 + 2) ^ (1 / 5)

ln(x^a)
=alnx
The derivative is a / X
ln(x^2 +2)^(1/5 )=1/5 × ln(x^2 +2)
The derivative is 2x / (5x ^ 2 + 10)

The derivative of ln|x|
.

When x > 0, it is obviously 1 / X
When x

Finding the indefinite integral of 1 / 1 + SiNx
Why can't it be reduced to 1 / (SiNx / 2 + cosx / 2) ^ 2 = 1 / (cosx / 2) ^ 2 * (TaNx / 2 + 1) ^ 2 = (secx / 2) ^ 2 / (TaNx / 2 + 1) ^ 2
His indefinite integral is 1 / (TaNx / 2 + 1)
Why is that not right

The one you reduced is not an elementary function. It can't be directly changed into DX ∫ 1 / (1 + SiNx) DX = ∫ [(1-sinx) / (1-sinx) (1 + SiNx)] DX = ∫ (1-sinx) / (1-sinx ^ 2 x) DX = ∫ [(1-sinx) / cos ^ 2 x] DX = ∫ [1 / cos ^ 2 x] DX + ∫ [1 / cos ^ 2 x] d (cosx) = ∫ [(sin ^ 2 x + co

[Online] given 1 + TaNx / 1-tanx = 3 + 2 √ 2, find the values of sinxcosx and SiNx · (sinx-3cosx),

TaNx = √ 2 / 2 can be calculated first, and then it can be calculated by the method of chord tangent
sinxcosx=sinxcosx/1=(sinxcosx)/(sin²x+cos²x)=tanx/(1+tan²x)=√2/3
The numerator and denominator are divided by cos & # 178; X
sinx·(sinx-3cosx)=(sin²x-3sinxcosx)/(sin²x+cos²x)=(tan²x-3tanx)/(1+tan²x)=(1-3√2)/3