Function y = 1 − LNX The derivative of 1 + LNX is () A. -2 (1+lnx)2 B. 2 x(1+lnx)2 C. -2 x(1+lnx)2 D. -1 x(1+lnx)2

Function y = 1 − LNX The derivative of 1 + LNX is () A. -2 (1+lnx)2 B. 2 x(1+lnx)2 C. -2 x(1+lnx)2 D. -1 x(1+lnx)2

By y = 1 − LNX
1+lnx，
So y '= (1 − LNX)
1+lnx)′＝(1−lnx)′(1+lnx)−(1−lnx)(1+lnx)′
(1+lnx)2=−2
x(1+lnx)2．
Therefore, C

What is the derivative of y = 1 + LNX / x? Find out the specific process,

[f(x)/g(x)]'=[f'(x)g(x)-f(x)g'(x)]/g(x)^2
So: y '= (LNX / x)' = [(LNX) '* x-lnx * x'] / x ^ 2 = [(1 / x) * x-lnx * 1] / x ^ 2 = (1-lnx) / x ^ 2
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Find the derivative of y = 2 ^ (x / LNX)? The reference answer is [2 ^ (x / LNX). LN2. (lnx-1)] / [(LNX) ^ 2]. I can't understand how it came from,

Pairs of pairs on both sides: LNY = (x / LNX) LN2
Derivative of X on both sides:
y'/y=(ln2)[lnx-x(1/x)]/ln ² x
=(ln2)(lnx-1)/ln ² x
Then: y '= (LN2) y (lnx-1) / LN ² x
=(ln2)2^(x/lnx)(lnx-1)/ln ² x
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Derivative of [ln (x + 1) - LNX] Why - 1 / X (x + 1)?

Solution 1:
[ln(x+1)-lnx]'
=[ln(x+1)]'-(lnx)'
=1/(x+1)-1/x
=x/[x(x+1)]-(x+1)/[x(x+1)]
=[x-(x+1)]/[x(x+1)]
=(x-x-1)/[x(x+1)]
=-1/[x(x+1)]

Find the derivative of y = ln (x ^ 1 / 2) + (LNX) ^ 1 / 2

Original formula = 1 / 2 * LNX + √ (LNX)
So y '= 1 / 2 * 1 / x + 1 / (2 √ LNX) * (LNX)'
=1/(2x)+1/(2√lnx)*1/x
=1/(2x)+1/(2x√lnx)

Find the derivative of the function y = ln √ x + √ LNX

Y'=1/√x *1/2√x+1/2√lnx* 1/x
=1/2x+1/(2x√lnx)