Solve 4/y-8+3/8-y

$$\frac { 4 } { y - 8 } + \frac { 3 } { 8 - y }$$

Evaluate

$$\frac{1}{y-8}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of y-8 and 8-y is y-8. Multiply \frac{3}{8-y} times \frac{-1}{-1}.

$$\frac{4}{y-8}+\frac{3\left(-1\right)}{y-8}$$

Since \frac{4}{y-8} and \frac{3\left(-1\right)}{y-8} have the same denominator, add them by adding their numerators.

$$\frac{4+3\left(-1\right)}{y-8}$$

Do the multiplications in 4+3\left(-1\right).

$$\frac{4-3}{y-8}$$

Do the calculations in 4-3.

$$\frac{1}{y-8}$$

Differentiate w.r.t. y

$$-\frac{1}{\left(y-8\right)^{2}}$$

Steps Using Chain Rule

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of y-8 and 8-y is y-8. Multiply \frac{3}{8-y} times \frac{-1}{-1}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{4}{y-8}+\frac{3\left(-1\right)}{y-8})$$

Since \frac{4}{y-8} and \frac{3\left(-1\right)}{y-8} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{4+3\left(-1\right)}{y-8})$$

Do the multiplications in 4+3\left(-1\right).

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{4-3}{y-8})$$

Do the calculations in 4-3.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{1}{y-8})$$

If F is the composition of two differentiable functions f\left(u\right) and u=g\left(x\right), that is, if F\left(x\right)=f\left(g\left(x\right)\right), then the derivative of F is the derivative of f with respect to u times the derivative of g with respect to x, that is, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

$$-\left(y^{1}-8\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}y}(y^{1}-8)$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$-\left(y^{1}-8\right)^{-2}y^{1-1}$$

Simplify.

$$-y^{0}\left(y^{1}-8\right)^{-2}$$

For any term t, t^{1}=t.

$$-y^{0}\left(y-8\right)^{-2}$$

For any term t except 0, t^{0}=1.

$$-\left(y-8\right)^{-2}$$