Solve d/dxleft(4x-3+1/x-5right)

$$\frac{d}{d x } \left(4x-3+ \frac{ 1 }{ x-5 } \right)$$

Evaluate

$$\frac{\left(2x-11\right)\left(2x-9\right)}{\left(x-5\right)^{2}}$$

Steps Using Derivative Rule for Quotient

To add or subtract expressions, expand them to make their denominators the same. Multiply 4x-3 times \frac{x-5}{x-5}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{\left(4x-3\right)\left(x-5\right)}{x-5}+\frac{1}{x-5})$$

Since \frac{\left(4x-3\right)\left(x-5\right)}{x-5} and \frac{1}{x-5} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{\left(4x-3\right)\left(x-5\right)+1}{x-5})$$

Do the multiplications in \left(4x-3\right)\left(x-5\right)+1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{4x^{2}-20x-3x+15+1}{x-5})$$

Combine like terms in 4x^{2}-20x-3x+15+1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{4x^{2}-23x+16}{x-5})$$

For any two differentiable functions, the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared.

$$\frac{\left(x^{1}-5\right)\frac{\mathrm{d}}{\mathrm{d}x}(4x^{2}-23x^{1}+16)-\left(4x^{2}-23x^{1}+16\right)\frac{\mathrm{d}}{\mathrm{d}x}(x^{1}-5)}{\left(x^{1}-5\right)^{2}}$$

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}.

$$\frac{\left(x^{1}-5\right)\left(2\times 4x^{2-1}-23x^{1-1}\right)-\left(4x^{2}-23x^{1}+16\right)x^{1-1}}{\left(x^{1}-5\right)^{2}}$$

Simplify.

$$\frac{\left(x^{1}-5\right)\left(8x^{1}-23x^{0}\right)-\left(4x^{2}-23x^{1}+16\right)x^{0}}{\left(x^{1}-5\right)^{2}}$$

Multiply x^{1}-5 times 8x^{1}-23x^{0}.

$$\frac{x^{1}\times 8x^{1}+x^{1}\left(-23\right)x^{0}-5\times 8x^{1}-5\left(-23\right)x^{0}-\left(4x^{2}-23x^{1}+16\right)x^{0}}{\left(x^{1}-5\right)^{2}}$$

Multiply 4x^{2}-23x^{1}+16 times x^{0}.

$$\frac{x^{1}\times 8x^{1}+x^{1}\left(-23\right)x^{0}-5\times 8x^{1}-5\left(-23\right)x^{0}-\left(4x^{2}x^{0}-23x^{1}x^{0}+16x^{0}\right)}{\left(x^{1}-5\right)^{2}}$$

To multiply powers of the same base, add their exponents.

$$\frac{8x^{1+1}-23x^{1}-5\times 8x^{1}-5\left(-23\right)x^{0}-\left(4x^{2}-23x^{1}+16x^{0}\right)}{\left(x^{1}-5\right)^{2}}$$

Simplify.

$$\frac{8x^{2}-23x^{1}-40x^{1}+115x^{0}-\left(4x^{2}-23x^{1}+16x^{0}\right)}{\left(x^{1}-5\right)^{2}}$$

Combine like terms.

$$\frac{4x^{2}-40x^{1}+99x^{0}}{\left(x^{1}-5\right)^{2}}$$

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

$$\frac{4x^{2}-40x+99x^{0}}{\left(x-5\right)^{2}}$$

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

$$\frac{4x^{2}-40x+99\times 1}{\left(x-5\right)^{2}}$$

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

$$\frac{4x^{2}-40x+99}{\left(x-5\right)^{2}}$$

Differentiate w.r.t. x

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