Differentiate y=4x+1/x?

Case I : y = (4x + 1)/x

y' = [x(4x - 1)' - (4x - 1)x'] / x²
= [x(4) - (4x - 1)] / x²
= (4x - 4x + 1) / x²
= 1/x²


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Case II : y = 4x + (1/x)

y' = (4x)' + (1/x)'
= 4 + (-1/x²)
= 4 - (1/x²)

Do you mean 4x + (1/x), or (4x+1)/x ?

4 - 1/x²

dy/dx = 4 - 1/x²

dy/dx = 4 - 1/x^2 for x <> 0

dy/dx = 4 - 1/x^2

Answer y =4x+1/x
dy/dx = d/dx(4x+1/x)
dy/dx = 4*1-1/x²
dy/dx = 4-1/x²

y'(x) = 4 - 1/x^2

dy/dx = 4 - 1/x^2

Well,

y = 4x + 1/x
then
dy/dx = 4 - 1/x^2

now, if you mean
y = (4x + 1)/x <---- then better PUT BRACKETS !! ;-)
dy/dx = ( 4x - (4x+1) ) /x^2
= -1/x^2

hope it' ll help !!

dy/dx=d(4x+1/x)/dx
D(y)= D(4x)+D(1/x)
D(y)=4+(-1/x^2)
D(y)=4-1/x^2

dy/dx = 4 - 1/x^2

dy/dx=4-1/x^2

4 - x^-2

y = 4x + x^(-1), dy/dx = 4 - (1/x^2)
y = (4x+1)/x = 4x/x + 1/x = 4+1/x , dy/dx = -1/x^2

y = 4x + x^(-1)
y' = 4 + (-1) x^(-2)
y' = 4 - 1 / x^2

dy/dx = 4 - 1/x²

y = 4x + 1/x
y = 4x + x^-1
dy/dx = 4 + (-1)(x^-2)
dy/dx = 4 - x^-2
dy/dx = 4 - 1/x^2

y= 4x+1/x
y= 4+1/x
y'= 0-1/x2
y'= -1/x2

-1/x^2