In the previous section, we completed a discussion on chain rule. We saw some solved examples also. In this section, we will see derivatives of implicit functions.
First we will see some basic details about explicit functions. It can be written in 7 steps:
1. Consider the equation: x−y−π = 0
2. This equation has two variables x and y. The equation can be rearranged and written as:
y = x − π
After the rearrangement, the equation has become a function. We see that, y is a function of x.
3. We can write either of the two ways:
♦ y = x − π
♦ f(x) = x − π
Now it is easy to find the derivative $\frac{dy}{dx}$ or f'(x).
4. So there are functions which can be easily rearranged so that three conditions are satisfied:
(i) There is a single "y" variable on the L.H.S.
(ii) All "x terms" are on the R.H.S.
(iii) There is no "y" variable or "y term" on the R.H.S.
•
Such functions are called explicit functions.
5. The equation xy = 1 is another example.
•
We can rearrange this equation and write: $y=\frac{1}{x}$
•
All three conditions that we wrote in (4) are satisfied.
•
So xy = 1 is an explicit function.
6.In the case of explicit functions, We say that:
y is given as an explicit function of x
7. The dictionary meaning of 'explicit' is:
Clear and exact. There will not be any need for doubts.
•
Indeed, in the case of explicit functions, we can clearly see how 'y' is dependent on 'x'.
Now we will see some basic details about implicit functions. It can be written in 5 steps:
1. Consider the equation: x+sin xy −y = 0
2. This equation has two variables x and y. The equation cannot be easily rearranged so as to satisfy the three conditions that we wrote in the case of explicit functions.
•
Such functions are called implicit functions.
3. In the case of implicit functions also, "y" is dependent on "x". But the manner of this dependency is not clear and exact.
•
We say that:
y is given as an implicit function of x.
That is., dependency of 'y' on 'x' is implied.
4. The dictionary meaning of 'implicit' is:
A fact is implied, but not communicated directly.
5. The equation x3y5 + 3x = 8y3 + 1
is another example.
•
We cannot rearrange it easily.
Let us see how to find the derivative in the case of implicit functions. It can be explained in steps:
So far in this chapter, we have been finding the derivatives of explicit functions.
Consider the explicit function f(x) = 2x+3.
•
This can be written as y = 2x+3 also.
•
Using 'y' instead of 'f(x)' is useful when we plot the graph of the function.
♦ We plot the input values along the x-axis.
♦ The output values are 'f(x) values'. They are plotted along the y-axis.
♦ So using 'y' instead of 'f(x)' will not make any difference.
•
We will now see a new method for finding the derivative. It can be explained using an example.
Let us first use an explicit function as example. The method can be written in 5 steps:
1. Consider the function x−y = π.
•
We want the derivative of this function with respect to x.
♦ That is., we want $\frac{dy}{dx}$.
2. Consider the "y" in the given equation. It is a function of "x" because, value of y depends on the value of x. So we can write "f(x)" in the place of "y". We get:
x−f(x) = π
(We need not worry about how f(x) is defined. This fact will become clear when we solve a few problems)
3. Differentiating both sides, we get:
$\frac{d}{dx} (x) \,-\, \frac{d}{dx} f(x) \,=\, \frac{d}{dx} (\pi)$
⇒ 1 − f'(x) = 0
⇒ f'(x) = 1
4. We wrote "f(x)" in the place of "y".
•
So f'(x) is $\frac{dy}{dx}$
•
Therefore, $\frac{dy}{dx}$ = 1
5. We will get the same result even if we begin with y = x−π.
Let us see another example to demonstrate the new method. This time also, we will use an explicit function.
1. Consider the function xy = 1.
•
We want the derivative of this function with respect to x.
♦ That is., we want $\frac{dy}{dx}$.
2.
Consider the "y" in the given equation. It is a function of "x"
because, value of y depends on the value of x. So we can write "f(x)" in
the place of "y". We get:
x f(x) = 1
(We need not worry about how f(x) is defined. This fact will become clear when we solve a few problems)
3. Differentiating both sides, we get:
$\frac{d}{dx} (x) . f(x) \,+\, x \frac{d}{dx} f(x) \,=\, \frac{d}{dx} (1)$
⇒ 1.f(x) + x f'(x) = 0
⇒ f'(x) = $-\frac{f(x)}{x}$
4. We wrote "f(x)" in the place of "y".
•
So f'(x) is $\frac{dy}{dx}$
•
Therefore, $\frac{dy}{dx} = -\frac{f(x)}{x} = - \frac{y}{x} = - \frac{1/x}{x} = -\frac{1}{x^2}$
5. We will get the same result even if we begin with $y=\frac{1}{x}$.
Now we will apply the new method to an implicit function.
1. Consider the function y + sin y = cos x.
•
We want the derivative of this function with respect to x.
♦ That is., we want $\frac{dy}{dx}$.
2.
Consider the "y" in the given equation. It is a function of "x"
because, value of y depends on the value of x. So we can write "f(x)" in
the place of "y". We get:
f(x) + sin (f(x)) = cos x
(We need not worry about how f(x) is defined. This fact will become clear when we solve a few problems)
3. Differentiating both sides, we get:
$\frac{d}{dx} (f(x)) \,+\, \frac{d}{dx} \sin(f(x)) \,=\, \frac{d}{dx} (\cos x)$
⇒ $f'(x) \,+\, \frac{d}{dx} \sin(f(x)) \,=\, - \sin x$
4. The second term in the L.H.S needs special attention. It is the derivative of a composite function. It can be determined in 4 steps:
(i) g(x) = sin (f(x)) = (v∘u)(x) = v(u(x))
Where u(x) = f(x) and v(u(x)) = sin (f(x)).
(ii) v'(u(x)) = cos (f(x))
(iii) u'(x) = f'(x)
(iv) So g'(x) = v'(u(x)).u'(x) = cos (f(x)).f'(x)
5. So the result in (3) becomes:
$f'(x) \,+\, \cos(f(x)).f'(x) \,=\, - \sin x$
⇒ $f'(x) \left[1 \,+\, \cos(f(x)) \right] \,=\, - \sin x$
⇒ $f'(x) \,=\, - \frac{\sin x}{1 \,+\, \cos(f(x))}$
6. We wrote "f(x)" in the place of "y".
•
So f'(x) is $\frac{dy}{dx}$
•
Therefore, $\frac{dy}{dx} = - \frac{\sin x}{1 \,+\, \cos(f(x))} = - \frac{\sin x}{1 \,+\, \cos y}$
•
Here, the denominator should not be zero. That means, cos y should not be −1.
•
So we can write: y ≠ (2n+1)π
In the next section, we will see some solved examples.
Copyright©2024 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment