Question

Find dy/dx, when x = te¹, y = 1 + log t.

Answer 1

Because, x = te¹

Taking logarithms of both the sides 

log x = log t + t log e = log t + t

Differentiating w.r.to 't'- (because log_e^e = 1)

or (d log/dx) x (dx/dt) = (d log t/dt) + (dt/dt) 

or, (1/x).(dx/dt) = (1/t) + 1