Calculus Examples

$x^{4} + t^{4} x + t^{3} - 9 = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d t} (x^{4} + t^{4} x + t^{3} - 9) = \frac{d}{d t} (0)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $x^{4} + t^{4} x + t^{3} - 9$ with respect to $t$ is $\frac{d}{d t} [x^{4}] + \frac{d}{d t} [t^{4} x] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 9]$ .

$\frac{d}{d t} [x^{4}] + \frac{d}{d t} [t^{4} x] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 9]$

Step 2.2

Evaluate $\frac{d}{d t} [x^{4}]$ .

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Step 2.2.1

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

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Step 2.2.1.1

To apply the Chain Rule, set $u$ as $x$ .

$\frac{d}{d u} [u^{4}] \frac{d}{d t} [x] + \frac{d}{d t} [t^{4} x] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 9]$

Step 2.2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 4$ .

$4 u^{3} \frac{d}{d t} [x] + \frac{d}{d t} [t^{4} x] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 9]$

Step 2.2.1.3

Replace all occurrences of $u$ with $x$ .

$4 x^{3} \frac{d}{d t} [x] + \frac{d}{d t} [t^{4} x] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 9]$

Step 2.2.2

Rewrite $\frac{d}{d t} [x]$ as $x'$ .

$4 x^{3} x' + \frac{d}{d t} [t^{4} x] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 9]$

Step 2.3

Evaluate $\frac{d}{d t} [t^{4} x]$ .

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Step 2.3.1

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 $f (t) = t^{4}$ and $g (t) = x$ .

$4 x^{3} x' + t^{4} \frac{d}{d t} [x] + x \frac{d}{d t} [t^{4}] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 9]$

Step 2.3.2

Rewrite $\frac{d}{d t} [x]$ as $x'$ .

$4 x^{3} x' + t^{4} x' + x \frac{d}{d t} [t^{4}] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 9]$

Step 2.3.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 4$ .

$4 x^{3} x' + t^{4} x' + x (4 t^{3}) + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 9]$

Step 2.3.4

Move $4$ to the left of $x$ .

$4 x^{3} x' + t^{4} x' + 4 x t^{3} + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 9]$

Step 2.4

Differentiate.

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Step 2.4.1

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 3$ .

$4 x^{3} x' + t^{4} x' + 4 x t^{3} + 3 t^{2} + \frac{d}{d t} [- 9]$

Step 2.4.2

Since $- 9$ is constant with respect to $t$ , the derivative of $- 9$ with respect to $t$ is $0$ .

$4 x^{3} x' + t^{4} x' + 4 x t^{3} + 3 t^{2} + 0$

Step 2.5

Simplify.

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Step 2.5.1

Add $4 x^{3} x' + t^{4} x' + 4 x t^{3} + 3 t^{2}$ and $0$ .

$4 x^{3} x' + t^{4} x' + 4 x t^{3} + 3 t^{2}$

Step 2.5.2

Reorder terms.

$3 t^{2} + 4 x^{3} x' + t^{4} x' + 4 t^{3} x$

Step 3

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

$0$

Step 4

Reform the equation by setting the left side equal to the right side.

$3 t^{2} + 4 x^{3} x' + t^{4} x' + 4 t^{3} x = 0$

Step 5

Solve for $x'$ .

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Step 5.1

Move all terms not containing $x'$ to the right side of the equation.

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Step 5.1.1

Subtract $3 t^{2}$ from both sides of the equation.

$4 x^{3} x' + t^{4} x' + 4 t^{3} x = - 3 t^{2}$

Step 5.1.2

Subtract $4 t^{3} x$ from both sides of the equation.

$4 x^{3} x' + t^{4} x' = - 3 t^{2} - 4 t^{3} x$

Step 5.2

Factor $x'$ out of $4 x^{3} x' + t^{4} x'$ .

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Step 5.2.1

Factor $x'$ out of $4 x^{3} x'$ .

$x' (4 x^{3}) + t^{4} x' = - 3 t^{2} - 4 t^{3} x$

Step 5.2.2

Factor $x'$ out of $t^{4} x'$ .

$x' (4 x^{3}) + x' (t^{4}) = - 3 t^{2} - 4 t^{3} x$

Step 5.2.3

Factor $x'$ out of $x' (4 x^{3}) + x' (t^{4})$ .

$x' (4 x^{3} + t^{4}) = - 3 t^{2} - 4 t^{3} x$

Step 5.3

Divide each term in $x' (4 x^{3} + t^{4}) = - 3 t^{2} - 4 t^{3} x$ by $4 x^{3} + t^{4}$ and simplify.

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Step 5.3.1

Divide each term in $x' (4 x^{3} + t^{4}) = - 3 t^{2} - 4 t^{3} x$ by $4 x^{3} + t^{4}$ .

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $4 x^{3} + t^{4}$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Divide $x'$ by $1$ .

$x' = \frac{- 3 t^{2}}{4 x^{3} + t^{4}} + \frac{- 4 t^{3} x}{4 x^{3} + t^{4}}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

$x' = \frac{- 3 t^{2} - 4 t^{3} x}{4 x^{3} + t^{4}}$

Step 5.3.3.2

Factor $t^{2}$ out of $- 3 t^{2} - 4 t^{3} x$ .

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Step 5.3.3.2.1

Factor $t^{2}$ out of $- 3 t^{2}$ .

$x' = \frac{t^{2} \cdot - 3 - 4 t^{3} x}{4 x^{3} + t^{4}}$

Step 5.3.3.2.2

Factor $t^{2}$ out of $- 4 t^{3} x$ .

$x' = \frac{t^{2} \cdot - 3 + t^{2} (- 4 t x)}{4 x^{3} + t^{4}}$

Step 5.3.3.2.3

Factor $t^{2}$ out of $t^{2} \cdot - 3 + t^{2} (- 4 t x)$ .

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

Step 5.3.3.3

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

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

Step 5.3.3.4

Factor $- 1$ out of $- 4 t x$ .

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

Step 5.3.3.5

Factor $- 1$ out of $- 1 (3) - (4 t x)$ .

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

Step 5.3.3.6

Move the negative in front of the fraction.

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

Step 6

Replace $x'$ with $\frac{d x}{d t}$ .

$\frac{d x}{d t} = - \frac{t^{2} (3 + 4 t x)}{4 x^{3} + t^{4}}$