# Derivative Of Parametric Functions

## The mathematics class, the topic of derivatives. Rules for finding derivatives of parametric functions. Lecturing and solved examples.

**Parametric Functions**

If the function y = f(x) has defined in the form as the following

y = g(x)

x = h(t)

t ϵ R,

The function y = f(x) is called parametric function.

Where the variable t is called parameter.

**Derivative Of Parametric Functions**

If the variable t can be eliminated easily in the parametric function, the variable t is eliminates and it is taken the derivative of obtained function.

If the variable t cannot eliminate, then it can be calculated the derivative of the parametric function using the chain rule.

y = g(x)

x = h(t)

h’(t) ≠ 0

dy |
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dx |

dy |
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dx |

**Example**

y = 3t^{6} + 2t

x = t^{3} + 4t

Find the dy/dx

**Solution:**

The functions x and y have shown in a t-dependent format.

We takes derivatives according to the "t" these functions . Then we ratio the functions and we can find the desired derivative.

y = 3t^{6} + 2t

y’ = 18t^{5} + 2

x = t^{3} + 4t

x’ = 3t^{2} + 4

dy |
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dx |

**Example:**

The functions x and y are defined as the following.

y = 3t^{6} – 15t^{2}

x = 2t^{6} – 10t^{2}

According to this,

A) Calculate the dy/dt

B) Calculate the dx/dt

C) Calculate the dy/dx

**Solution:**

A)

The dy/dt is y‘(t)

y’(t) = 18t^{5} – 30t

B)

The dx/dy is x’(t)

x’(t) = 12t^{5}– 20t

C)

The dy/dx = y’(x)

dy |
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dx |

= | 18t^{5} – 30t | |

12t^{5} – 20t |

= | 6(3t^{5} – 5t) | |

4(t^{5} – 5t) |

= | 6 | |

4 |

= | 3 | |

2 |

**Example:**

y =6t + 5

x =√ 3t – 4

What is the value of f’(x) for x = 5?

**Solution:**

In this example the parameter t can be eliminated easily. We will eliminate the parameter t and we obtain the function y = f(x).

y =6t + 5

x = √3t – 4

Let we take the square of the x

x^{2} = 3t – 4

Now, the system of equation is as the following.

y = 6t + 5

x^{2} = 3t – 4

Let we multiplication the second equation by 2.

y = 6t + 5

-2x^{2} = 8 – 6t

Now, let we do addition the two equations reciprocal.

y – 2x^{2} = 13

y = 2x^{2} + 13

We obtained the function y = f(x). Now we can be calculate the derivative of the y.

y’ = 4x

For x = 5,

f’(5) = 4.5 = 20

**Example:**

y = 4u^{5} – 5u

x = 5u^{4} + 10u

Calculate the dy/dx

Solution:

dy/du = 20u^{4} – 5

dx/du = 20u^{4}+ 10

= | dy |
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dx |

= | 4u^{4} – 1 | |

4u^{4} + 2 |

**Example**

The functions t and x are defined as the following.

t = 8u^{3} + 6u^{2}

x = 5u^{4} – 2u

Calculate the value of dt/dx for u = 1.

**Solution:**

Let we find the derivative of each functions.

dt/du = 24u^{2} + 12u

dx/du = 20u^{3} – 2

Let we ratio these two derivatives.

dt |
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dx |

For x = 1

dt |
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dx |

= 2

**Example:**

x = e^{5t}

y = e^{2t}

Calculate the dy/dx

**Solution:**

Let we find the derivative of functions x and y.

x’ = 5e^{5t}

y’ = 2e^{2t}

Now, We will ratio these derivatives.

dy |
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dx |

= 2e^{(2t – 5t)/5}

= 2e^{-3t/5}

**Derivative Of The Sum Of Functions**

**Derivative Of The Quotient Of Functions**

RISE KNOWLEDGE

March 8 2019

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