Solve 3/x^2+2x-15:15/x-3=

$$\frac { 3 } { x ^ { 2 } + 2 x - 15 } : \frac { 15 } { x - 3 } =$$

Evaluate

$$\frac{1}{5\left(x+5\right)}$$

Short Solution Steps

Divide \frac{3}{x^{2}+2x-15} by \frac{15}{x-3} by multiplying \frac{3}{x^{2}+2x-15} by the reciprocal of \frac{15}{x-3}.

$$\frac{3\left(x-3\right)}{\left(x^{2}+2x-15\right)\times 15}$$

Cancel out 3 in both numerator and denominator.

$$\frac{x-3}{5\left(x^{2}+2x-15\right)}$$

Factor the expressions that are not already factored.

$$\frac{x-3}{5\left(x-3\right)\left(x+5\right)}$$

Cancel out x-3 in both numerator and denominator.

$$\frac{1}{5\left(x+5\right)}$$

Expand the expression.

$$\frac{1}{5x+25}$$

Differentiate w.r.t. x

$$-\frac{1}{5\left(x+5\right)^{2}}$$

Steps Using Chain Rule

Divide \frac{3}{x^{2}+2x-15} by \frac{15}{x-3} by multiplying \frac{3}{x^{2}+2x-15} by the reciprocal of \frac{15}{x-3}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3\left(x-3\right)}{\left(x^{2}+2x-15\right)\times 15})$$

Cancel out 3 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x-3}{5\left(x^{2}+2x-15\right)})$$

Factor the expressions that are not already factored in \frac{x-3}{5\left(x^{2}+2x-15\right)}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x-3}{5\left(x-3\right)\left(x+5\right)})$$

Cancel out x-3 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{5\left(x+5\right)})$$

Use the distributive property to multiply 5 by x+5.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{5x+25})$$

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).

$$-\left(5x^{1}+25\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}x}(5x^{1}+25)$$

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}.

$$-\left(5x^{1}+25\right)^{-2}\times 5x^{1-1}$$

Simplify.

$$-5x^{0}\left(5x^{1}+25\right)^{-2}$$

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

$$-5x^{0}\left(5x+25\right)^{-2}$$

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

$$-5\left(5x+25\right)^{-2}$$