Calculus Examples

$t \sqrt{4 - t}$

Find the first derivative.

Tap for more steps...

Find the first derivative.

Tap for more steps...

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 - t}$ as ${(4 - t)}^{\frac{1}{2}}$ .

$f' (t) = \frac{d}{d t} (t {(4 - t)}^{\frac{1}{2}})$

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = t$ and $g (t) = {(4 - t)}^{\frac{1}{2}}$ .

$f' (t) = t \frac{d}{d t} ({(4 - t)}^{\frac{1}{2}}) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{\frac{1}{2}}$ and $g (t) = 4 - t$ .

Tap for more steps...

To apply the Chain Rule, set $u$ as $4 - t$ .

$f' (t) = t (\frac{d}{d u} (u^{\frac{1}{2}}) \frac{d}{d t} (4 - t)) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{1}{2}$ .

$f' (t) = t (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d t} (4 - t)) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Replace all occurrences of $u$ with $4 - t$ .

$f' (t) = t (\frac{1}{2} \cdot {(4 - t)}^{\frac{1}{2} - 1} \frac{d}{d t} (4 - t)) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (t) = t (\frac{1}{2} \cdot {(4 - t)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d t} (4 - t)) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Combine $- 1$ and $\frac{2}{2}$ .

$f' (t) = t (\frac{1}{2} \cdot {(4 - t)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d t} (4 - t)) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Combine the numerators over the common denominator.

$f' (t) = t (\frac{1}{2} \cdot {(4 - t)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d t} (4 - t)) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Simplify the numerator.

Tap for more steps...

Multiply $- 1$ by $2$ .

$f' (t) = t (\frac{1}{2} \cdot {(4 - t)}^{\frac{1 - 2}{2}} \frac{d}{d t} (4 - t)) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Subtract $2$ from $1$ .

$f' (t) = t (\frac{1}{2} \cdot {(4 - t)}^{\frac{- 1}{2}} \frac{d}{d t} (4 - t)) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Combine fractions.

Tap for more steps...

Move the negative in front of the fraction.

$f' (t) = t (\frac{1}{2} \cdot {(4 - t)}^{- \frac{1}{2}} \frac{d}{d t} (4 - t)) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Combine $\frac{1}{2}$ and ${(4 - t)}^{- \frac{1}{2}}$ .

$f' (t) = t (\frac{{(4 - t)}^{- \frac{1}{2}}}{2} \cdot \frac{d}{d t} (4 - t)) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Move ${(4 - t)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (t) = t (\frac{1}{2 {(4 - t)}^{\frac{1}{2}}} \cdot \frac{d}{d t} (4 - t)) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Combine $\frac{1}{2 {(4 - t)}^{\frac{1}{2}}}$ and $t$ .

$f' (t) = \frac{t}{2 {(4 - t)}^{\frac{1}{2}}} \cdot \frac{d}{d t} (4 - t) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

By the Sum Rule, the derivative of $4 - t$ with respect to $t$ is $\frac{d}{d t} [4] + \frac{d}{d t} [- t]$ .

$f' (t) = \frac{t}{2 {(4 - t)}^{\frac{1}{2}}} \cdot (\frac{d}{d t} (4) + \frac{d}{d t} (- t)) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Since $4$ is constant with respect to $t$ , the derivative of $4$ with respect to $t$ is $0$ .

$f' (t) = \frac{t}{2 {(4 - t)}^{\frac{1}{2}}} \cdot (0 + \frac{d}{d t} (- t)) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Add $0$ and $\frac{d}{d t} [- t]$ .

$f' (t) = \frac{t}{2 {(4 - t)}^{\frac{1}{2}}} \cdot \frac{d}{d t} (- t) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Since $- 1$ is constant with respect to $t$ , the derivative of $- t$ with respect to $t$ is $- \frac{d}{d t} [t]$ .

$f' (t) = \frac{t}{2 {(4 - t)}^{\frac{1}{2}}} \cdot (- \frac{d}{d t} t) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$f' (t) = \frac{t}{2 {(4 - t)}^{\frac{1}{2}}} \cdot (- 1 \cdot 1) + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Combine fractions.

Tap for more steps...

Multiply $- 1$ by $1$ .

$f' (t) = \frac{t}{2 {(4 - t)}^{\frac{1}{2}}} \cdot - 1 + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Combine $\frac{t}{2 {(4 - t)}^{\frac{1}{2}}}$ and $- 1$ .

$f' (t) = \frac{t \cdot - 1}{2 {(4 - t)}^{\frac{1}{2}}} + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Simplify the expression.

Tap for more steps...

Move $- 1$ to the left of $t$ .

$f' (t) = \frac{- 1 \cdot t}{2 {(4 - t)}^{\frac{1}{2}}} + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Rewrite $- 1 t$ as $- t$ .

$f' (t) = \frac{- t}{2 {(4 - t)}^{\frac{1}{2}}} + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Move the negative in front of the fraction.

$f' (t) = - \frac{t}{2 {(4 - t)}^{\frac{1}{2}}} + {(4 - t)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$f' (t) = - \frac{t}{2 {(4 - t)}^{\frac{1}{2}}} + {(4 - t)}^{\frac{1}{2}} \cdot 1$

Multiply ${(4 - t)}^{\frac{1}{2}}$ by $1$ .

$f' (t) = - \frac{t}{2 {(4 - t)}^{\frac{1}{2}}} + {(4 - t)}^{\frac{1}{2}}$

To write ${(4 - t)}^{\frac{1}{2}}$ as a fraction with a common denominator, multiply by $\frac{2 {(4 - t)}^{\frac{1}{2}}}{2 {(4 - t)}^{\frac{1}{2}}}$ .

$f' (t) = - \frac{t}{2 {(4 - t)}^{\frac{1}{2}}} + {(4 - t)}^{\frac{1}{2}} \cdot \frac{2 {(4 - t)}^{\frac{1}{2}}}{2 {(4 - t)}^{\frac{1}{2}}}$

Combine ${(4 - t)}^{\frac{1}{2}}$ and $\frac{2 {(4 - t)}^{\frac{1}{2}}}{2 {(4 - t)}^{\frac{1}{2}}}$ .

$f' (t) = - \frac{t}{2 {(4 - t)}^{\frac{1}{2}}} + \frac{{(4 - t)}^{\frac{1}{2}} (2 {(4 - t)}^{\frac{1}{2}})}{2 {(4 - t)}^{\frac{1}{2}}}$

Combine the numerators over the common denominator.

$f' (t) = \frac{- t + {(4 - t)}^{\frac{1}{2}} (2 {(4 - t)}^{\frac{1}{2}})}{2 {(4 - t)}^{\frac{1}{2}}}$

Multiply ${(4 - t)}^{\frac{1}{2}}$ by ${(4 - t)}^{\frac{1}{2}}$ by adding the exponents.

Tap for more steps...

Move ${(4 - t)}^{\frac{1}{2}}$ .

$f' (t) = \frac{- t + {(4 - t)}^{\frac{1}{2}} {(4 - t)}^{\frac{1}{2}} \cdot 2}{2 {(4 - t)}^{\frac{1}{2}}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f' (t) = \frac{- t + {(4 - t)}^{\frac{1}{2} + \frac{1}{2}} \cdot 2}{2 {(4 - t)}^{\frac{1}{2}}}$

Combine the numerators over the common denominator.

$f' (t) = \frac{- t + {(4 - t)}^{\frac{1 + 1}{2}} \cdot 2}{2 {(4 - t)}^{\frac{1}{2}}}$

Add $1$ and $1$ .

$f' (t) = \frac{- t + {(4 - t)}^{\frac{2}{2}} \cdot 2}{2 {(4 - t)}^{\frac{1}{2}}}$

Divide $2$ by $2$ .

$f' (t) = \frac{- t + (4 - t) \cdot 2}{2 {(4 - t)}^{\frac{1}{2}}}$

Simplify ${(4 - t)}^{1} \cdot 2$ .

$f' (t) = \frac{- t + (4 - t) \cdot 2}{2 {(4 - t)}^{\frac{1}{2}}}$

Move $2$ to the left of $4 - t$ .

$f' (t) = \frac{- t + 2 \cdot (4 - t)}{2 {(4 - t)}^{\frac{1}{2}}}$

Simplify.

Tap for more steps...

Apply the distributive property.

$f' (t) = \frac{- t + 2 \cdot 4 + 2 (- t)}{2 {(4 - t)}^{\frac{1}{2}}}$

Simplify the numerator.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $2$ by $4$ .

$f' (t) = \frac{- t + 8 + 2 (- t)}{2 {(4 - t)}^{\frac{1}{2}}}$

Multiply $- 1$ by $2$ .

$f' (t) = \frac{- t + 8 - 2 t}{2 {(4 - t)}^{\frac{1}{2}}}$

Subtract $2 t$ from $- t$ .

$f' (t) = \frac{- 3 t + 8}{2 {(4 - t)}^{\frac{1}{2}}}$

Factor $- 1$ out of $- 3 t$ .

$f' (t) = \frac{- (3 t) + 8}{2 {(4 - t)}^{\frac{1}{2}}}$

Rewrite $8$ as $- 1 (- 8)$ .

$f' (t) = \frac{- (3 t) - 1 \cdot - 8}{2 {(4 - t)}^{\frac{1}{2}}}$

Factor $- 1$ out of $- (3 t) - 1 (- 8)$ .

$f' (t) = \frac{- (3 t - 8)}{2 {(4 - t)}^{\frac{1}{2}}}$

Rewrite $- (3 t - 8)$ as $- 1 (3 t - 8)$ .

$f' (t) = \frac{- 1 (3 t - 8)}{2 {(4 - t)}^{\frac{1}{2}}}$

Move the negative in front of the fraction.

$f' (t) = - \frac{3 t - 8}{2 {(4 - t)}^{\frac{1}{2}}}$

The first derivative of $f (t)$ with respect to $t$ is $- \frac{3 t - 8}{2 {(4 - t)}^{\frac{1}{2}}}$ .

$- \frac{3 t - 8}{2 {(4 - t)}^{\frac{1}{2}}}$

Set the first derivative equal to $0$ then solve the equation $- \frac{3 t - 8}{2 {(4 - t)}^{\frac{1}{2}}} = 0$ .

Tap for more steps...

Set the first derivative equal to $0$ .

$- \frac{3 t - 8}{2 {(4 - t)}^{\frac{1}{2}}} = 0$

Set the numerator equal to zero.

$3 t - 8 = 0$

Solve the equation for $t$ .

Tap for more steps...

Add $8$ to both sides of the equation.

$3 t = 8$

Divide each term in $3 t = 8$ by $3$ and simplify.

Tap for more steps...

Divide each term in $3 t = 8$ by $3$ .

$\frac{3 t}{3} = \frac{8}{3}$

Simplify the left side.

Tap for more steps...

Cancel the common factor of $3$ .

Tap for more steps...

Cancel the common factor.

$\frac{3 t}{3} = \frac{8}{3}$

Divide $t$ by $1$ .

$t = \frac{8}{3}$

Find the values where the derivative is undefined.

Tap for more steps...

Convert expressions with fractional exponents to radicals.

Tap for more steps...

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$- \frac{3 t - 8}{2 \sqrt{{(4 - t)}^{1}}}$

Anything raised to $1$ is the base itself.

$- \frac{3 t - 8}{2 \sqrt{4 - t}}$

Set the denominator in $\frac{3 t - 8}{2 \sqrt{4 - t}}$ equal to $0$ to find where the expression is undefined.

$2 \sqrt{4 - t} = 0$

Solve for $t$ .

Tap for more steps...

To remove the radical on the left side of the equation, square both sides of the equation.

${(2 \sqrt{4 - t})}^{2} = 0^{2}$

Simplify each side of the equation.

Tap for more steps...

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 - t}$ as ${(4 - t)}^{\frac{1}{2}}$ .

${(2 {(4 - t)}^{\frac{1}{2}})}^{2} = 0^{2}$

Simplify the left side.

Tap for more steps...

Simplify ${(2 {(4 - t)}^{\frac{1}{2}})}^{2}$ .

Tap for more steps...

Apply the product rule to $2 {(4 - t)}^{\frac{1}{2}}$ .

$2^{2} {({(4 - t)}^{\frac{1}{2}})}^{2} = 0^{2}$

Raise $2$ to the power of $2$ .

$4 {({(4 - t)}^{\frac{1}{2}})}^{2} = 0^{2}$

Multiply the exponents in ${({(4 - t)}^{\frac{1}{2}})}^{2}$ .

Tap for more steps...

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 {(4 - t)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Cancel the common factor of $2$ .

Tap for more steps...

Cancel the common factor.

$4 {(4 - t)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Rewrite the expression.

$4 {(4 - t)}^{1} = 0^{2}$

Simplify.

$4 (4 - t) = 0^{2}$

Apply the distributive property.

$4 \cdot 4 + 4 (- t) = 0^{2}$

Multiply.

Tap for more steps...

Multiply $4$ by $4$ .

$16 + 4 (- t) = 0^{2}$

Multiply $- 1$ by $4$ .

$16 - 4 t = 0^{2}$

Simplify the right side.

Tap for more steps...

Raising $0$ to any positive power yields $0$ .

$16 - 4 t = 0$

Solve for $t$ .

Tap for more steps...

Subtract $16$ from both sides of the equation.

$- 4 t = - 16$

Divide each term in $- 4 t = - 16$ by $- 4$ and simplify.

Tap for more steps...

Divide each term in $- 4 t = - 16$ by $- 4$ .

$\frac{- 4 t}{- 4} = \frac{- 16}{- 4}$

Simplify the left side.

Tap for more steps...

Cancel the common factor of $- 4$ .

Tap for more steps...

Cancel the common factor.

$\frac{- 4 t}{- 4} = \frac{- 16}{- 4}$

Divide $t$ by $1$ .

$t = \frac{- 16}{- 4}$

Simplify the right side.

Tap for more steps...

Divide $- 16$ by $- 4$ .

$t = 4$

Set the radicand in $\sqrt{4 - t}$ less than $0$ to find where the expression is undefined.

$4 - t < 0$

Solve for $t$ .

Tap for more steps...

Subtract $4$ from both sides of the inequality.

$- t < - 4$

Divide each term in $- t < - 4$ by $- 1$ and simplify.

Tap for more steps...

Divide each term in $- t < - 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- t}{- 1} > \frac{- 4}{- 1}$

Simplify the left side.

Tap for more steps...

Dividing two negative values results in a positive value.

$\frac{t}{1} > \frac{- 4}{- 1}$

Divide $t$ by $1$ .

$t > \frac{- 4}{- 1}$

Simplify the right side.

Tap for more steps...

Divide $- 4$ by $- 1$ .

$t > 4$

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$t \geq 4$

$[4, \infty)$

$t \geq 4$

$[4, \infty)$

Evaluate $t \sqrt{4 - t}$ at each $t$ value where the derivative is $0$ or undefined.

Tap for more steps...

Evaluate at $t = \frac{8}{3}$ .

Tap for more steps...

Substitute $\frac{8}{3}$ for $t$ .

$(\frac{8}{3}) \sqrt{4 - (\frac{8}{3})}$

Simplify.

Tap for more steps...

To write $4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{8}{3} \sqrt{4 \cdot \frac{3}{3} - \frac{8}{3}}$

Combine $4$ and $\frac{3}{3}$ .

$\frac{8}{3} \sqrt{\frac{4 \cdot 3}{3} - \frac{8}{3}}$

Combine the numerators over the common denominator.

$\frac{8}{3} \sqrt{\frac{4 \cdot 3 - 8}{3}}$

Simplify the numerator.

Tap for more steps...

Multiply $4$ by $3$ .

$\frac{8}{3} \sqrt{\frac{12 - 8}{3}}$

Subtract $8$ from $12$ .

$\frac{8}{3} \sqrt{\frac{4}{3}}$

Rewrite $\sqrt{\frac{4}{3}}$ as $\frac{\sqrt{4}}{\sqrt{3}}$ .

$\frac{8}{3} \cdot \frac{\sqrt{4}}{\sqrt{3}}$

Simplify the numerator.

Tap for more steps...

Rewrite $4$ as $2^{2}$ .

$\frac{8}{3} \cdot \frac{\sqrt{2^{2}}}{\sqrt{3}}$

Pull terms out from under the radical, assuming positive real numbers.

$\frac{8}{3} \cdot \frac{2}{\sqrt{3}}$

Multiply $\frac{2}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$\frac{8}{3} (\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}})$

Combine and simplify the denominator.

Tap for more steps...

Multiply $\frac{2}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$\frac{8}{3} \cdot \frac{2 \sqrt{3}}{\sqrt{3} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$\frac{8}{3} \cdot \frac{2 \sqrt{3}}{{\sqrt{3}}^{1} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$\frac{8}{3} \cdot \frac{2 \sqrt{3}}{{\sqrt{3}}^{1} {\sqrt{3}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{8}{3} \cdot \frac{2 \sqrt{3}}{{\sqrt{3}}^{1 + 1}}$

Add $1$ and $1$ .

$\frac{8}{3} \cdot \frac{2 \sqrt{3}}{{\sqrt{3}}^{2}}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

Tap for more steps...

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$\frac{8}{3} \cdot \frac{2 \sqrt{3}}{{(3^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{8}{3} \cdot \frac{2 \sqrt{3}}{3^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$\frac{8}{3} \cdot \frac{2 \sqrt{3}}{3^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Tap for more steps...

Cancel the common factor.

$\frac{8}{3} \cdot \frac{2 \sqrt{3}}{3^{\frac{2}{2}}}$

Rewrite the expression.

$\frac{8}{3} \cdot \frac{2 \sqrt{3}}{3^{1}}$

Evaluate the exponent.

$\frac{8}{3} \cdot \frac{2 \sqrt{3}}{3}$

Multiply $\frac{8}{3} \cdot \frac{2 \sqrt{3}}{3}$ .

Tap for more steps...

Multiply $\frac{8}{3}$ by $\frac{2 \sqrt{3}}{3}$ .

$\frac{8 (2 \sqrt{3})}{3 \cdot 3}$

Multiply $2$ by $8$ .

$\frac{16 \sqrt{3}}{3 \cdot 3}$

Multiply $3$ by $3$ .

$\frac{16 \sqrt{3}}{9}$

Evaluate at $t = 4$ .

Tap for more steps...

Substitute $4$ for $t$ .

$(4) \sqrt{4 - (4)}$

Simplify.

Tap for more steps...

Multiply $- 1$ by $4$ .

$4 \sqrt{4 - 4}$

Subtract $4$ from $4$ .

$4 \sqrt{0}$

Rewrite $0$ as $0^{2}$ .

$4 \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$4 \cdot 0$

Multiply $4$ by $0$ .

$0$

List all of the points.

$(\frac{8}{3}, \frac{16 \sqrt{3}}{9}), (4, 0)$