Test: Logarithmic Functions - JEE MCQ

 d(x^y)=d(e^(y⋅log(x)))
=e^(y⋅log(x))d(y⋅log(x))
=(x^y)(dy⋅log(x) + y⋅d(log(x))
=(x^y) 
d(y^x) = (y^x)(log(y)dx + x/ydy). 
Since  d(x^y) = d(y^x) , and simplifying by  x^y = y^x , we get
log(y)dx + x/ydy = log(x)dy + y/xdx. 
Removing the denominators leads to:
xylog(y)dx + x²dy = xylog(x)dy + y²dx 
(xylog(y) − y²)dx = (xylog(x) − x²)dy 
dy/dx = (xylog(y)−y²)/(xy⋅log(x)−x²)

dy/dx = 2^x(tanx)' + tanx (2^x)'
dy/dx = 2^x (secx)² + 2^x tanx log2 , {since , (a^x)' = a^x loga (x)'}
dy/dx = 2^x[(Secx)² + log2 tanx]

y = (lnx)^tanx
ln(y) = tanx(ln(lnx))
d(lny)/dx = d(tanx(ln(lnx)))/dx
Using product rule -
(1/y)dy/dx = (secx)²(ln(lnx)) + (1÷xlnx)tanx
dy/dx = [(secx)²(ln(lnx)) + (1÷xlnx)tanx ]×y
dy/dx = [(secx)²(ln(lnx)) + (1÷xlnx)tanx ]×[(lnx)^tanx]

 y = (1 - log x)/(1 + log x)
Applying the Quoitent rule and differential for ln x
dy / dx = [(1 - ln x)(1 / x) - (1+ ln x)( -1 / x)]/(1 - ln x)²
= [(1 - ln x + 1 + ln x)]/x(1 + ln x)²
= 2 / x(1+ln x)²

Given equation,

Taking log both the sides
xlogy = (x-y)loge
Differentiate it with respect to x, we get
x/y dy/dx = logy = loge - xloge dy/dx 
dy/dx = (loge - logy)/(x/y + loge)  
= y(1 - loge)/(x + y)

On differentiating both sides with respect to x