Calculus Examples

$x^{2} + t^{2} x + t^{3} - 5 = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d t} (x^{2} + t^{2} x + t^{3} - 5) = \frac{d}{d t} (0)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $x^{2} + t^{2} x + t^{3} - 5$ with respect to $t$ is $\frac{d}{d t} [x^{2}] + \frac{d}{d t} [t^{2} x] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 5]$ .

$\frac{d}{d t} [x^{2}] + \frac{d}{d t} [t^{2} x] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 5]$

Step 2.2

Evaluate $\frac{d}{d t} [x^{2}]$ .

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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 $f (t) = t^{2}$ and $g (t) = x$ .

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Step 2.2.1.1

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

$\frac{d}{d u} [u^{2}] \frac{d}{d t} [x] + \frac{d}{d t} [t^{2} x] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 5]$

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 $n = 2$ .

$2 u \frac{d}{d t} [x] + \frac{d}{d t} [t^{2} x] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 5]$

Step 2.2.1.3

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

$2 x \frac{d}{d t} [x] + \frac{d}{d t} [t^{2} x] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 5]$

Step 2.2.2

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

$2 x x' + \frac{d}{d t} [t^{2} x] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 5]$

Step 2.3

Evaluate $\frac{d}{d t} [t^{2} x]$ .

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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 $f (t) = t^{2}$ and $g (t) = x$ .

$2 x x' + t^{2} \frac{d}{d t} [x] + x \frac{d}{d t} [t^{2}] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 5]$

Step 2.3.2

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

$2 x x' + t^{2} x' + x \frac{d}{d t} [t^{2}] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 5]$

Step 2.3.3

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

$2 x x' + t^{2} x' + x (2 t) + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 5]$

Step 2.3.4

Move $2$ to the left of $x$ .

$2 x x' + t^{2} x' + 2 x t + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 5]$

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 $n = 3$ .

$2 x x' + t^{2} x' + 2 x t + 3 t^{2} + \frac{d}{d t} [- 5]$

Step 2.4.2

Since $- 5$ is constant with respect to $t$ , the derivative of $- 5$ with respect to $t$ is $0$ .

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

Step 2.5

Simplify.

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Step 2.5.1

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

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

Step 2.5.2

Reorder terms.

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

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.

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

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.

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

Step 5.1.2

Subtract $2 t x$ from both sides of the equation.

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

Step 5.2

Factor $x'$ out of $2 x x' + t^{2} x'$ .

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Step 5.2.1

Factor $x'$ out of $2 x x'$ .

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

Step 5.2.2

Factor $x'$ out of $t^{2} x'$ .

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

Step 5.2.3

Factor $x'$ out of $x' (2 x) + x' (t^{2})$ .

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $2 x + t^{2}$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Divide $x'$ by $1$ .

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

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

Step 5.3.3.2

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

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Step 5.3.3.2.1

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

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

Step 5.3.3.2.2

Factor $t$ out of $- 2 t x$ .

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

Step 5.3.3.2.3

Factor $t$ out of $t (- 3 t) + t (- 2 x)$ .

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

Step 5.3.3.3

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

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

Step 5.3.3.4

Factor $- 1$ out of $- 2 x$ .

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

Step 5.3.3.5

Factor $- 1$ out of $- (3 t) - (2 x)$ .

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

Step 5.3.3.6

Simplify the expression.

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Step 5.3.3.6.1

Rewrite $- (3 t + 2 x)$ as $- 1 (3 t + 2 x)$ .

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

Step 5.3.3.6.2

Move the negative in front of the fraction.

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

Step 6

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

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