Solve =4t/frac{1{t}-1/5}

$$= \frac { 4 t } { \frac { 1 } { t } - \frac { 1 } { 5 } }$$

Evaluate

$$-\frac{20t^{2}}{t-5}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of t and 5 is 5t. Multiply \frac{1}{t} times \frac{5}{5}. Multiply \frac{1}{5} times \frac{t}{t}.

$$\frac{4t}{\frac{5}{5t}-\frac{t}{5t}}$$

Since \frac{5}{5t} and \frac{t}{5t} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4t}{\frac{5-t}{5t}}$$

Divide 4t by \frac{5-t}{5t} by multiplying 4t by the reciprocal of \frac{5-t}{5t}.

$$\frac{4t\times 5t}{5-t}$$

Multiply t and t to get t^{2}.

$$\frac{4t^{2}\times 5}{5-t}$$

Multiply 4 and 5 to get 20.

$$\frac{20t^{2}}{5-t}$$

Differentiate w.r.t. t

$$\frac{20t\left(10-t\right)}{\left(t-5\right)^{2}}$$

Steps Using Derivative Rule for Quotient

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of t and 5 is 5t. Multiply \frac{1}{t} times \frac{5}{5}. Multiply \frac{1}{5} times \frac{t}{t}.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{4t}{\frac{5}{5t}-\frac{t}{5t}})$$

Since \frac{5}{5t} and \frac{t}{5t} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{4t}{\frac{5-t}{5t}})$$

Divide 4t by \frac{5-t}{5t} by multiplying 4t by the reciprocal of \frac{5-t}{5t}.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{4t\times 5t}{5-t})$$

Multiply t and t to get t^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{4t^{2}\times 5}{5-t})$$

Multiply 4 and 5 to get 20.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{20t^{2}}{5-t})$$

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.

$$\frac{\left(-t^{1}+5\right)\frac{\mathrm{d}}{\mathrm{d}t}(20t^{2})-20t^{2}\frac{\mathrm{d}}{\mathrm{d}t}(-t^{1}+5)}{\left(-t^{1}+5\right)^{2}}$$

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}.

$$\frac{\left(-t^{1}+5\right)\times 2\times 20t^{2-1}-20t^{2}\left(-1\right)t^{1-1}}{\left(-t^{1}+5\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(-t^{1}+5\right)\times 40t^{1}-20t^{2}\left(-1\right)t^{0}}{\left(-t^{1}+5\right)^{2}}$$

Expand using distributive property.

$$\frac{-t^{1}\times 40t^{1}+5\times 40t^{1}-20t^{2}\left(-1\right)t^{0}}{\left(-t^{1}+5\right)^{2}}$$

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

$$\frac{-40t^{1+1}+5\times 40t^{1}-20\left(-1\right)t^{2}}{\left(-t^{1}+5\right)^{2}}$$

Do the arithmetic.

$$\frac{-40t^{2}+200t^{1}-\left(-20t^{2}\right)}{\left(-t^{1}+5\right)^{2}}$$

Combine like terms.

$$\frac{\left(-40-\left(-20\right)\right)t^{2}+200t^{1}}{\left(-t^{1}+5\right)^{2}}$$

Subtract -20 from -40.

$$\frac{-20t^{2}+200t^{1}}{\left(-t^{1}+5\right)^{2}}$$

Factor out 20t.

$$\frac{20t\left(-t^{1}+10t^{0}\right)}{\left(-t^{1}+5\right)^{2}}$$

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

$$\frac{20t\left(-t+10t^{0}\right)}{\left(-t+5\right)^{2}}$$

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

$$\frac{20t\left(-t+10\times 1\right)}{\left(-t+5\right)^{2}}$$

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

$$\frac{20t\left(-t+10\right)}{\left(-t+5\right)^{2}}$$