Solve 1/t-1+3/3t-1

$$\frac { 1 } { t - 1 } + \frac { 3 } { 3 t - 1 }$$

Evaluate

$$\frac{2\left(3t-2\right)}{\left(t-1\right)\left(3t-1\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of t-1 and 3t-1 is \left(t-1\right)\left(3t-1\right). Multiply \frac{1}{t-1} times \frac{3t-1}{3t-1}. Multiply \frac{3}{3t-1} times \frac{t-1}{t-1}.

$$\frac{3t-1}{\left(t-1\right)\left(3t-1\right)}+\frac{3\left(t-1\right)}{\left(t-1\right)\left(3t-1\right)}$$

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

$$\frac{3t-1+3\left(t-1\right)}{\left(t-1\right)\left(3t-1\right)}$$

Do the multiplications in 3t-1+3\left(t-1\right).

$$\frac{3t-1+3t-3}{\left(t-1\right)\left(3t-1\right)}$$

Combine like terms in 3t-1+3t-3.

$$\frac{6t-4}{\left(t-1\right)\left(3t-1\right)}$$

Expand \left(t-1\right)\left(3t-1\right).

$$\frac{6t-4}{3t^{2}-4t+1}$$

Differentiate w.r.t. t

$$\frac{2\left(-9t^{2}+12t-5\right)}{\left(\left(t-1\right)\left(3t-1\right)\right)^{2}}$$

Steps Using Derivative Rule for Quotient

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of t-1 and 3t-1 is \left(t-1\right)\left(3t-1\right). Multiply \frac{1}{t-1} times \frac{3t-1}{3t-1}. Multiply \frac{3}{3t-1} times \frac{t-1}{t-1}.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{3t-1}{\left(t-1\right)\left(3t-1\right)}+\frac{3\left(t-1\right)}{\left(t-1\right)\left(3t-1\right)})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{3t-1+3\left(t-1\right)}{\left(t-1\right)\left(3t-1\right)})$$

Do the multiplications in 3t-1+3\left(t-1\right).

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{3t-1+3t-3}{\left(t-1\right)\left(3t-1\right)})$$

Combine like terms in 3t-1+3t-3.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{6t-4}{\left(t-1\right)\left(3t-1\right)})$$

Apply the distributive property by multiplying each term of t-1 by each term of 3t-1.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{6t-4}{3t^{2}-t-3t+1})$$

Combine -t and -3t to get -4t.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{6t-4}{3t^{2}-4t+1})$$

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.

$$\frac{\left(3t^{2}-4t^{1}+1\right)\frac{\mathrm{d}}{\mathrm{d}t}(6t^{1}-4)-\left(6t^{1}-4\right)\frac{\mathrm{d}}{\mathrm{d}t}(3t^{2}-4t^{1}+1)}{\left(3t^{2}-4t^{1}+1\right)^{2}}$$

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}.

$$\frac{\left(3t^{2}-4t^{1}+1\right)\times 6t^{1-1}-\left(6t^{1}-4\right)\left(2\times 3t^{2-1}-4t^{1-1}\right)}{\left(3t^{2}-4t^{1}+1\right)^{2}}$$

Simplify.

$$\frac{\left(3t^{2}-4t^{1}+1\right)\times 6t^{0}-\left(6t^{1}-4\right)\left(6t^{1}-4t^{0}\right)}{\left(3t^{2}-4t^{1}+1\right)^{2}}$$

Multiply 3t^{2}-4t^{1}+1 times 6t^{0}.

$$\frac{3t^{2}\times 6t^{0}-4t^{1}\times 6t^{0}+6t^{0}-\left(6t^{1}-4\right)\left(6t^{1}-4t^{0}\right)}{\left(3t^{2}-4t^{1}+1\right)^{2}}$$

Multiply 6t^{1}-4 times 6t^{1}-4t^{0}.

$$\frac{3t^{2}\times 6t^{0}-4t^{1}\times 6t^{0}+6t^{0}-\left(6t^{1}\times 6t^{1}+6t^{1}\left(-4\right)t^{0}-4\times 6t^{1}-4\left(-4\right)t^{0}\right)}{\left(3t^{2}-4t^{1}+1\right)^{2}}$$

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

$$\frac{3\times 6t^{2}-4\times 6t^{1}+6t^{0}-\left(6\times 6t^{1+1}+6\left(-4\right)t^{1}-4\times 6t^{1}-4\left(-4\right)t^{0}\right)}{\left(3t^{2}-4t^{1}+1\right)^{2}}$$

Simplify.

$$\frac{18t^{2}-24t^{1}+6t^{0}-\left(36t^{2}-24t^{1}-24t^{1}+16t^{0}\right)}{\left(3t^{2}-4t^{1}+1\right)^{2}}$$

Combine like terms.

$$\frac{-18t^{2}+24t^{1}-10t^{0}}{\left(3t^{2}-4t^{1}+1\right)^{2}}$$

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

$$\frac{-18t^{2}+24t-10t^{0}}{\left(3t^{2}-4t+1\right)^{2}}$$

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

$$\frac{-18t^{2}+24t-10}{\left(3t^{2}-4t+1\right)^{2}}$$