Solve 1/4-3x+x^2

$$\frac{ 1 }{ 4-3x+ { x }^{ 2 } }$$

Differentiate w.r.t. x

$$-\frac{2x-3}{\left(x^{2}-3x+4\right)^{2}}$$

Steps Using Chain Rule

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).

$$-\left(-3x^{1}+x^{2}+4\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}x}(-3x^{1}+x^{2}+4)$$

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}.

$$-\left(-3x^{1}+x^{2}+4\right)^{-2}\left(-3x^{1-1}+2x^{2-1}\right)$$

Simplify.

$$\left(-3x^{1}+x^{2}+4\right)^{-2}\left(3x^{0}-2x^{1}\right)$$

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

$$\left(-3x+x^{2}+4\right)^{-2}\left(3x^{0}-2x\right)$$

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

$$\left(-3x+x^{2}+4\right)^{-2}\left(3\times 1-2x\right)$$

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

$$\left(-3x+x^{2}+4\right)^{-2}\left(3-2x\right)$$

Evaluate

$$\frac{1}{x^{2}-3x+4}$$