Solve 1/x^2+10x-24-0/5x-10

$$\frac { 1 } { x ^ { 2 } + 10 x - 24 } - \frac { 0 } { 5 x - 10 }$$

Evaluate

$$\frac{1}{\left(x-2\right)\left(x+12\right)}$$

Short Solution Steps

Zero divided by any non-zero term gives zero.

$$\frac{1}{x^{2}+10x-24}+0$$

Factor x^{2}+10x-24.

$$\frac{1}{\left(x-2\right)\left(x+12\right)}+0$$

Anything plus zero gives itself.

$$\frac{1}{x^{2}+10x-24}$$

Differentiate w.r.t. x

$$\frac{2\left(-x-5\right)}{\left(\left(x-2\right)\left(x+12\right)\right)^{2}}$$

Steps Using Chain Rule

Zero divided by any non-zero term gives zero.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x^{2}+10x-24}+0)$$

Factor x^{2}+10x-24.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{\left(x-2\right)\left(x+12\right)}+0)$$

Anything plus zero gives itself.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x^{2}+10x-24})$$

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).

$$-\left(x^{2}+10x^{1}-24\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}+10x^{1}-24)$$

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}.

$$-\left(x^{2}+10x^{1}-24\right)^{-2}\left(2x^{2-1}+10x^{1-1}\right)$$

Simplify.

$$\left(x^{2}+10x^{1}-24\right)^{-2}\left(-2x^{1}-10x^{0}\right)$$

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

$$\left(x^{2}+10x-24\right)^{-2}\left(-2x-10x^{0}\right)$$

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

$$\left(x^{2}+10x-24\right)^{-2}\left(-2x-10\right)$$