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 
=dy 
dt
dx
dt
dx





dy
g’(t)
h’(t)
dx




Example

y = 3t6 + 2t

x = t3 + 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 = 3t6 + 2t

y’ = 18t5 + 2


x = t3 + 4t

x’ = 3t2 + 4

dy 
=18t5 + 2
3t2 + 4
dx




Example:

The functions x and y are defined as the following.

y = 3t6 – 15t2

x = 2t6 – 10t2

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) = 18t5 – 30t


B) 

The dx/dy is x’(t)

x’(t) = 12t5– 20t


C) 

The dy/dx = y’(x)

dy 
=dy
.dt
dx
dt
dx




=18t5 – 30t
12t5 – 20t




=6(3t5 – 5t)
4(t5 – 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 

x2 = 3t – 4 


Now, the system of equation is as the following.

y = 6t + 5

x2 = 3t – 4


Let we multiplication the second equation by 2.

y = 6t + 5

-2x2 = 8 – 6t

Now, let we do addition the two equations reciprocal.

y – 2x2 = 13

y = 2x2 + 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 = 4u5 – 5u

x = 5u4 + 10u

Calculate the dy/dx 


Solution:

dy/du = 20u4 – 5 

dx/du = 20u4+ 10


=dy 
=20u4 – 5
20u4 + 10
dx




=4u4 – 1
4u4 + 2




Example

The functions t and x are defined as the following.

t = 8u3 + 6u2

x = 5u4 – 2u

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


Solution:

Let we find the derivative of each functions.

dt/du = 24u2 + 12u

dx/du = 20u3 – 2


Let we ratio these two derivatives.

dt 
=24u2 + 12u
20u3 – 2
dx




For x = 1


dt 
(1) =36
18
dx




= 2


Example:

x = e5t

y = e2t

Calculate the dy/dx


Solution:

Let we find the derivative of functions x and y.

x’ = 5e5t

y’ = 2e2t


Now, We will ratio these derivatives.

dy 
=2e2t
5e5t
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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