the chain rule of differentiation
Remember,
$${[f(g(x))]’ = f’(g(x)) \cdot g’(x)}$$
or in Leibniz notation,
$${\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\cdot \frac{\mathrm{d}u}{\mathrm{d}x}}$$
which can also be written as
$${\frac{\mathrm{d}y}{\mathrm{d}u}=\frac{\mathrm{d}y}{\mathrm{d}x}\cdot \frac{\mathrm{d}x}{\mathrm{d}u}}$$
task
If
$${\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x}{\sqrt{x^2+1}}}$$
and ${u = x^2 + 1}$
, write down an expression for ${\frac{\mathrm{d}y}{\mathrm{d}x}}$
and hence find ${\frac{\mathrm{d}y}{\mathrm{d}u}}$
(a) in terms of ${x}$
, (b) in terms of ${u}$
.
solution
example
If $${\frac{\mathrm{d}y}{\mathrm{d}x}=x^2(2x+3)^4}$$ and ${u = 2x + 3}$
, find terms ${\frac{\mathrm{d}y}{\mathrm{d}u}}$
in terms of ${u}$
.
solution
example 1
Gesucht ist das unbestimmte Integral $${\int{(4x + 1)^3,\mathrm{d}x}}$$
1. Substitution
$${4x + 1 = z}$$
2. Differentiale
$${z^\prime = \frac{\mathrm{d}z}{\mathrm{d}x} =4}$$ $${\Rightarrow \mathrm{d}x = \frac{\mathrm{d}z}{4}}$$
3. Einsetzen von 1. und 2. ins Integral
$${\begin{align} \int{(4x + 1)^3,\mathrm{d}x} & = \int{z^3,\frac{\mathrm{d}z}{4}} \newline & = \frac{1}{16} z^4 + \mathrm{C}\end{align}}$$
4. Resubstitution
$${z = 4x + 1}$$ $${\begin{align} \int{(4x + 1)^3,\mathrm{d}x} = \frac{1}{16} (4x + 1)^4 + \mathrm{C}\end{align}}$$
example 2
Find $${y = \int{x \cos{(3x^2 + 5)},\mathrm{d}x}}$$
solution
$${z = 3x^2 + 5}$$ 2. $${z^\prime = \frac{\mathrm{d}z}{\mathrm{d}x} = 6x}$$ $${\Rightarrow \mathrm{d}x = \frac{\mathrm{d}z}{6x}}$$ 3. $${\begin{align} \int{x \cos{(3x^2 + 5)},\mathrm{d}x} & = \int{x \cos{(z)},\frac{\mathrm{d}z}{6x}} \newline & = \int{\cos{(z)},\frac{\mathrm{d}z}{6}} \newline & = \frac{1}{6} \sin{(z)} + \mathrm{C} \newline & = \frac{1}{6} \sin{(3x^2 + 5)} + \mathrm{C} \end{align}}$$
definite integrals
What happens to the upper and lower limit of the integrals?
If you substitute back after integrating, you keep the initial limits of the integral. And, if you decide not to substitute back after integrating, you have to adjust the limits by substituting them both into the epression for z (or u or whatever the new variable is called)
example 3 - difinite integral
Evaluate $${\int \limits_{0}^{2}{\frac{4x}{\sqrt{1 + 2x^2}}},\mathrm{d}x}$$
solution
method
First select the “inner function z” and the “outer function” so that
- the derivative z’ (or
${\frac{\mathrm{d}z}{\mathrm{d}x}}$
) is a factor of the function in the integral. - the anti-derivative of the “outer function” is known.
-
substitution $${z = 1 + 2x^2}$$
-
derivative (differentiale) $${z’ = \frac{\mathrm{d}z}{\mathrm{d}x} = 4x}$$ $${\Rightarrow \mathrm{d}x = \frac{\mathrm{d}z}{4x}}$$ $${\Rightarrow \mathrm{d}z = 4x \mathrm{d}z}$$
-
converting the limits
-
lower limit:
${z = 1 + 2\cdot 0^2 = 1}$
-
upper limit:
${z = 1 + 2\cdot 2^2 = 9}$
-
-
integration
${\int \limits_{0}^{2}{\frac{4x}{\sqrt{1 + 2x^2}}}\,\mathrm{d}x = \int \limits_{0}^{2}{\frac{1}{\sqrt{1 + 2x^2}} \cdot 4x}\,\mathrm{d}x = \int \limits_{1}^{9}{\frac{1}{\sqrt{z}}}\,\mathrm{d}z = \left[2 \sqrt {z} \right]_{1}^{9} = 4}$
Note You can resubstitute ${z = 1 + 2x^2}$
which would lead to the same result.
$${\int \limits_{0}^{2}{\frac{4x}{\sqrt{1 + 2x^2}}},\mathrm{d}x = … =\left[2 \sqrt {z} \right]_{1}^{9} = \left[2 \sqrt {1 + 2x^2} \right]_{0}^{2} = 4}$$
example 4
Find the integral of
$${\int {\frac{x}{\sqrt{1 - x}}},\mathrm{d}x}$$ by substituting $z=\sqrt{1-x}$
Solution - click on image to enlarge.
example 5 - Using trigonometry and Pythagoras
Evaluate
$${\int \limits_{0}^{\frac{1}{2}}{\frac{1}{\sqrt{1 - x^2}}},\mathrm{d}x}$$
remember that $${\sin^2{x} + \cos^2{x} = 1}$$ $${\Rightarrow \cos{x} = \sqrt{1 - \sin^2{x}}}$$
So, we
- substitute
${x = \sin{t}}$
- find the derivative
${x' = \frac{\mathrm{d}x}{\mathrm{d}t} = \cos(t)}$
${\Rightarrow \mathrm{d}x = \cos(t) \mathrm{d}t}$
- converting the limits
- lower limit:
${\sin(t) = 0 \Rightarrow t_1 = 0}$
- upper limit:
${\sin(t) = \frac{1}{2} \Rightarrow t_2 = \frac{\pi}{6}}$
- lower limit:
- integration
${\int \limits_{0}^{\frac{1}{2}}{\frac{1}{\sqrt{1 - x^2}}}\,\mathrm{d}x = \int \limits_{0}^{\frac{\pi}{6}}{\frac{1}{\sqrt{1 - (\sin(t)^2)}}} \cdot \cos(t)\,\mathrm{d}t = \int \limits_{0}^{\frac{\pi}{6}}{\frac{1}{\cos(t)}} \cdot \cos(t)\,\mathrm{d}t = \int \limits_{0}^{\frac{\pi}{6}}{1}\,\mathrm{d}t =\left[t \right]_{0}^{\frac{\pi}{6}} = \frac{\pi}{6}}$