Solve 1/41/x^2-x-4

$$\frac { 1 } { 4 } \frac { 1 } { x ^ { 2 } - x - 4 }$$

Evaluate

$$\frac{1}{4\left(x^{2}-x-4\right)}$$

Solution Steps

Multiply \frac{1}{4} times \frac{1}{x^{2}-x-4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{4\left(x^{2}-x-4\right)}$$

Use the distributive property to multiply 4 by x^{2}-x-4.

$$\frac{1}{4x^{2}-4x-16}$$

Differentiate w.r.t. x

$$-\frac{2x-1}{4\left(x^{2}-x-4\right)^{2}}$$

Steps Using Chain Rule

Multiply \frac{1}{4} times \frac{1}{x^{2}-x-4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{4\left(x^{2}-x-4\right)})$$

Use the distributive property to multiply 4 by x^{2}-x-4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{4x^{2}-4x-16})$$

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).

$$-\left(4x^{2}-4x^{1}-16\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}x}(4x^{2}-4x^{1}-16)$$

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}.

$$-\left(4x^{2}-4x^{1}-16\right)^{-2}\left(2\times 4x^{2-1}-4x^{1-1}\right)$$

Simplify.

$$\left(4x^{2}-4x^{1}-16\right)^{-2}\left(-8x^{1}+4x^{0}\right)$$

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

$$\left(4x^{2}-4x-16\right)^{-2}\left(-8x+4x^{0}\right)$$

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

$$\left(4x^{2}-4x-16\right)^{-2}\left(-8x+4\times 1\right)$$

For any term t, t\times 1=t and 1t=t.

$$\left(4x^{2}-4x-16\right)^{-2}\left(-8x+4\right)$$