In the previous section, we saw the basics about chain rule. We saw some solved examples also. In this section, we will see an alternate method for applying this rule.
The alternate method can be explained in 6 steps:
1. Recall the first example that we saw in the previous section.
• If f(x) = (3x+4)3, then
$\frac{df}{dx} \,=\, 9 (3x+4)^2$
2. The above result can be split up as:
$\frac{df}{dx} \,=\, 9 (3x+4)^2\,=\,\left[3 (3x+4)^2 \right].3$
So the result has two parts:
(i) $\left[3 (3x+4)^2 \right]$
(ii) 3
3. The function in our example is: f(x) = (3x+4)3.
We can write it as the composite of two functions: f(x) = v(u(x))
• Where
(i) u(x) = 3x+4
(ii) v(u(x)) = [u(x)]3.
• Then f(x) = v(u(x)) = v(3x+4) = (3x+4)3.
4. Consider 3(ii) above. From that equation, we get:
$v'(u(x))\,=\,3[u(x)]^2\,=\,3 (3x+4)^2$
• So part (i) of the result that we wrote in step 2 is v'(u(x)).
5. Consider 3(i) above. From that equation, we get:
$u'(x)\,=\,3$
• So part (ii) of the result that we wrote in step 2 is u'(x).
6. Based on the above steps, we can write the procedure in brief:
(i) We are given f(x) = v(u(x)). We want f'(x).
(ii) First consider u(x) as a whole, and differentiate it. That means, we are finding v'(u(x))
(iii) Then differentiate u(x). That means, we are finding u'(x)
(iv) Finally, multiply the results in (ii) and (iii).
6. If it is a composite of 3 functions, then also, we can write the procedure in brief:
(i) We are given f(x) =w(v(u(x))). We want f'(x).
(ii) First consider v(u(x)) as a whole, and differentiate it. That means, we are finding w'(v(u(x)))
(iii) Next consider u(x) as a whole, and differentiate it. That means, we are finding v'(u(x))
(iv) Next differentiate u(x). That means, we are finding u'(x)
(v) Finally, multiply the results in (ii), (iii) and (iv).
Let us revisit the solved examples that we saw in the previous section.
Solved example 21.21
Find the derivative of the function given by f(x) = sin (x2)
Solution (alternate method):
1. f(x) = sin (x2) = (v∘u)(x) = v(u(x))
Where u(x) = x2 and v(u(x)) = sin (u(x)).
2. v'(u(x)) = cos (x2)
3. u'(x) = 2x
4. So f'(x) = v'(u(x)).u'(x) = 2x cos (x2)
Solved example 21.22
Find the derivative of the function given by f(x) = tan (2x+3)
Solution (alternate method):
1. f(x) = tan (2x+3) = (v∘u)(x) = v(u(x))
Where u(x) = 2x+3 and v(u(x)) = tan (u(x)).
2. v'(u(x)) = sec2(2x+3)
3. u'(x) = 2
4. So f'(x) = v'(u(x)).u'(x) = 2 sec2(2x+3)
Solved example 21.23
Find the derivative of the function given by f(x) = sin (cos (x2))
Solution (alternate method):
1. f(x) = sin (cos (x2)) = (w∘v∘u)(x) = (w∘v)(u(x))
Where u(x) = x2, v(u(x)) = cos (u(x)) and w(v(u(x))) = sin (v(u(x))).
2. w'(v(u(x))) = cos (cos (x2))
3. v'(u(x)) = −sin (x2)
4. u'(x) = 2x
5. So f'(x) = w'(v(u(x))).v'(u(x)).u'(x)
= −2x sin (x2) cos (cos (x2))
Solved example 21.24
Find the derivative of the function given by
$f(x) = \sqrt{5x - 8}$
Solution (alternate method):
1. $f(x) = \sqrt{5x - 8}$ = (v∘u)(x) = v(u(x))
Where u(x) = 5x −8 and v(u(x)) = $\sqrt{u(x)}$.
2. v'(u(x)) = $\frac{1}{2} (5x - 8)^{- \frac{1}{2}}$
3. u'(x) = 5
4. So f'(x) = v'(u(x)).u'(x) = $\frac{5}{2 \sqrt{5x - 8}}$
Now we will see some new solved examples:
Solved example 21.25
Find the derivative of the function given by
$f(x) = \sin \left(3x^2 + x \right)$
Solution:
1. $f(x) = \sin \left(3x^2 + x \right)$ = (v∘u)(x) = v(u(x))
Where u(x) = (3x2 + x) and v(u(x)) = sin (u(x)).
2. v'(u(x)) = cos (3x2 + x)
3. u'(x) = 6x + 1
4. So f'(x) = v'(u(x)).u'(x) = (6x+1) cos (3x2 + x)
Solved example 21.26
Find the derivative of the function given by
f(x) = sec (1 − 5x)
Solution:
1. f(x) = sec (1 − 5x) = (v∘u)(x) = v(u(x))
Where u(x) = (1 − 5x) and v(u(x)) = sec (u(x)).
2. v'(u(x)) = sec (1 − 5x) tan (1 − 5x)
3. u'(x) = −5
4. So f'(x) = v'(u(x)).u'(x) = −5sec (1 − 5x) tan (1 − 5x)
Solved example 21.27
Find the derivative of the function given by
$f(x) = \left(2 x^3 \,+\, \cos x \right)^{50}$
Solution:
1. $f(x) = \left(2 x^3 \,+\, \cos x \right)^{50}$ = (v∘u)(x) = v(u(x))
Where $u(x) =\left(2 x^3 \,+\, \cos x \right)$ and $v(u(x)) = (u(x))^{50}$.
2. v'(u(x)) = $50\left(2 x^3 \,+\, \cos x \right)^{49}$
3. To find u'(x):
Part (i): differentiating 2x3:
The answer is 6x2.
Part (ii): differentiating cos x:
The answer is −sin x.
• So u'(x) = 6x2 − sin x.
4. So $f'(x) = 50 \left(6x^2 - \sin x \right) \left(2 x^3 \,+\, \cos x \right)^{49}$
Solved example 21.28
Find the derivative of the function given by
f(x) = cos2 (x)
Solution:
1. f(x) = cos2 (x) = (v∘u)(x) = v(u(x))
Where u(x) = cos (x) and v(u(x)) = (u(x))2 .
2. v'(u(x)) = 2 u(x) = 2 cos (x)
3. u'(x) = −sin (x)
4. So f'(x) = v'(u(x)).u'(x) = −2 sin (x) cos (x)
(This result can be obtained using product rule also)
Solved example 21.29
Find the derivative of the function given by
f(x) = sin2 (x)
Solution:
1. f(x) = sin2 (x) = (v∘u)(x) = v(u(x))
Where u(x) = sin (x) and v(u(x)) = (u(x))2 .
2. v'(u(x)) = 2 u(x) = 2 sin (x)
3. u'(x) = cos (x)
4. So f'(x) = v'(u(x)).u'(x) = 2 sin (x) cos (x)
(This result can be obtained using product rule also)
Solved example 21.30
Find the derivative of the function given by
f(x) = cos4 (x) + cos (x4)
Solution:
• The given function is of the form:
f(x) = g(x) + h(x)
• Applying algebra of derivatives, we can write:
f'(x) = g'(x) + h'(x)
Part (i): Finding g'(x)
1. g(x) = cos4 (x) = (v∘u)(x) = v(u(x))
Where u(x) = cos2 (x) and v(u(x)) = (u(x))2 .
2. v'(u(x)) = 2 u(x) = 2 cos2 (x)
3. u'(x) = −2 sin (x) cos (x)
4. So f'(x) = v'(u(x)).u'(x) = −4 sin (x) cos3 (x)
Part (ii): Finding h'(x)
1. h(x) = cos (x4) = (v∘u)(x) = v(u(x))
Where u(x) = x4 and v(u(x)) = cos (u(x)) .
2. v'(u(x)) = −sin (u(x)) = −sin (x4)
3. u'(x) = 4x3
4. So f'(x) = v'(u(x)).u'(x) = −4x3 sin (x4)
◼ Using the results from parts (i) and (ii), we can write:
f'(x) = −4 sin (x) cos3 (x) −4x3 sin (x4)
Solved example 21.31
Find the derivative of the function given by
f(x) = [g(x)]n
Solution:
1. f(x) = [g(x)]n = (v∘u)(x) = v(u(x))
Where u(x) = g(x) and v(u(x)) = (u(x))n .
2. v'(u(x)) = n(u(x))(n-1) = n(g(x))(n-1)
3. u'(x) = g'(x)
4. So f'(x) = v'(u(x)).u'(x) = n[g(x)](n-1) g'(x).
• The above result gives us an easy method to find the derivative in those cases where f(x) is the power of another function.
•
For example, consider f(x) = cos2x.
Here, g(x) = cos x and n = 2.
So f'(x) = 2[cos x]. −sin x = −2 sin x cos x.
Link to a few more solved examples is given below:
We have completed a discussion on chain rule. In the next section, we will see derivatives of implicit functions.
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