Solve 3x^0/y+2y^-1-y^-3/y^-2

$$\frac { 3 x ^ { 0 } } { y } + 2 y ^ { - 1 } - \frac { y ^ { - 3 } } { y ^ { - 2 } }$$

Evaluate

$$\frac{4}{y}$$

Short Solution Steps

Rewrite y^{-2} as y^{-3}y. Cancel out y^{-3} in both numerator and denominator.

$$\frac{3x^{0}}{y}+2y^{-1}-\frac{1}{y}$$

Calculate x to the power of 0 and get 1.

$$\frac{3\times 1}{y}+2y^{-1}-\frac{1}{y}$$

Multiply 3 and 1 to get 3.

$$\frac{3}{y}+2y^{-1}-\frac{1}{y}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2y^{-1} times \frac{y}{y}.

$$\frac{3}{y}+\frac{2y^{-1}y}{y}-\frac{1}{y}$$

Since \frac{3}{y} and \frac{2y^{-1}y}{y} have the same denominator, add them by adding their numerators.

$$\frac{3+2y^{-1}y}{y}-\frac{1}{y}$$

Do the multiplications in 3+2y^{-1}y.

$$\frac{3+2}{y}-\frac{1}{y}$$

Do the calculations in 3+2.

$$\frac{5}{y}-\frac{1}{y}$$

Since \frac{5}{y} and \frac{1}{y} have the same denominator, subtract them by subtracting their numerators. Subtract 1 from 5 to get 4.

$$\frac{4}{y}$$

Differentiate w.r.t. y

$$-\frac{4}{y^{2}}$$

Steps Using Definition of a Derivative

Rewrite y^{-2} as y^{-3}y. Cancel out y^{-3} in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{3x^{0}}{y}+2y^{-1}-\frac{1}{y})$$

Calculate x to the power of 0 and get 1.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{3\times 1}{y}+2y^{-1}-\frac{1}{y})$$

Multiply 3 and 1 to get 3.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{3}{y}+2y^{-1}-\frac{1}{y})$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2y^{-1} times \frac{y}{y}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{3}{y}+\frac{2y^{-1}y}{y}-\frac{1}{y})$$

Since \frac{3}{y} and \frac{2y^{-1}y}{y} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{3+2y^{-1}y}{y}-\frac{1}{y})$$

Do the multiplications in 3+2y^{-1}y.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{3+2}{y}-\frac{1}{y})$$

Do the calculations in 3+2.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{5}{y}-\frac{1}{y})$$

Since \frac{5}{y} and \frac{1}{y} have the same denominator, subtract them by subtracting their numerators. Subtract 1 from 5 to get 4.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{4}{y})$$

The derivative of ax^{n} is nax^{n-1}.

$$-4y^{-1-1}$$

Subtract 1 from -1.

$$-4y^{-2}$$