Question

√x+logx differentiate with respect to x​

Answer 1

$\huge\mathfrak\orange{Answer}$

y = √x+logx

So, dy/dx = d(√x)/dx + d(logx)/dx

dy/dx = 1/(2√x) + 1/x

So, dy/dx = dy/dx = 1/(2√x) + 1/x.