Methods of Differentiation
The method of determining a derivative is termed differentiation. Normally a dependent variable is represented in terms of independent variables utilising an equation. Below listed are the various methods of differentiation that we will discuss with proper explanation and related formulas in the coming headings.
- Differentiation using Chain Rule
- Differentiation using Product Rule
- Differentiation using Quotient Rule
- Differentiation through Logarithm
- Differentiation of Parametric Functions
- Differentiation of Implicit Functions
Differentiation using Chain Rule
The chain rule is applied when you have to locate the derivative of the composition of two functions. In this method, first, the derivative of the outer function is taken, then it is multiplied by the derivative of the inner function. The related is as follows:
\(\frac{d}{dx}(f(g(x)))=f’(g(x)).g’(x)\text{ or}\ \frac{dy}{dx}=\frac{dy}{dt}.\ \frac{dt}{dx}\)
Differentiation using Product Rule
Product Rule is used to find the derivative of the product of two functions. Herein the first function is multiplied with the derivative of the second + the second function is multiplied with the derivative of the first. The product rule formula for two and three variable is as shown:
\(\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\)
\(\frac{d}{dx}(uvw)=uv\frac{d(w)}{dx}+uw\frac{d(v)}{dx}+vw\frac{d(u)}{dx}\)
Also learn the various Applications of Derivatives here.
Differentiation using Quotient Rule
The quotient rule(division rule) or method is helpful when you need to find the derivative of a function that is in p/q format. The formula for the quotient rule for two variables is as follows:
\(\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\left(\frac{du}{dx}\right)-u\left(\frac{dv}{dx}\right)}{v^2}\)
Learn about First Principles of Derivatives and Derivative of Sec Square x
Differentiation through Logarithm
To simplify the differentiation of some functions we incorporate logarithm followed by differentiation. The logarithm concept is applied in two cases;
- When the function is a product of some other function, then the log transforms the product into a sum and then differentiation is carried out.
- When some variable occurs in the exponent of the function.
Derivative of \( u^v \) where u, v are differentiable functions of x.
Consider \( y = u^v \)
Take log on both sides
\(\log_{ }y=v.\ \log_{ }u\\\)
Next, differentiate w.r.t x
\(\frac{1}{y}.\ \frac{dy}{dx}=\frac{d}{dx}\left(v.\ \log_{ }u\right)\\
\frac{dy}{dx}=y.\ \frac{d}{dx}\left(v.\ \log_{ }u\right)=u^v.\ \frac{d}{dx}\left(v.\ \log_{ }u\right)\\ \)
Example: Find \(\frac{dy}{dx}\text{ for }y=(4x+2)^x\).
Solution:
Given:\( y=(4x+2)^x\).
Step 1: Taking logarithm of both the sides
\(\log y=x.\log(4x+2)\)
Step 2: Differentiating both sides with respect to x
\(\left(\frac{1}{y}\right)\ \frac{dy}{dx}=\left(\frac{4x}{(4x+2)}\right)+\log\left(4x+2\right)\) [Using product rule]
\(\frac{dy}{dx}=y.\left[\left(\frac{4x}{(4x+2)}\right)+\log\left(4x+2\right)\right]\)
Hence,
\(\frac{dy}{dx}=(4x+2)^x.\left[\left(\frac{4x}{(4x+2)}\right)+\log\left(4x+2\right)\right]\).
Learn methods of Solving Linear Differential Equations
Differentiation of Parametric Functions
If x and y are two variables, such that they are explicitly expressed in terms of another variable consider t i.e x = f(t), y = g(t), then these functions are said to be parametric functions where t(the third variable) is the parameter, then;
\(\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{g^{^{\prime}}\left(t\right)}{f^{^{\prime}}\left(t\right)}\ and\ \ \frac{dx}{dy}=\frac{\frac{dx}{dt}}{\frac{dy}{dt}}=\frac{f^{^{\prime}}\left(t\right)}{g^{^{\prime}}\left(t\right)}\)
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Differentiation of Implicit Functions
If an equation has both x and y together in an equation like f(x, y) = 0 then x (or y) is named the implicit function of y (or x). To solve such an equation:
Each term of f (x, y) = 0 should be differentiated with respect to x.
Next, keep the terms having dy/dx on one side the rest on the other side.
Compose the value of dy/dx as a function of x or y or both.
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Differentiation by Substitution
These are some important substitutions that helps to reduce the inverse trigonometric functions involving the terms as given below in the table while finding the derivatives of such functions:
| S. No. | Function involving terms | Substitutions |
| 1 | \(a^2 – x^2\) | x = a sin θ or x = a cos θ |
| 2 | \(a^2 + x^2\) | x = a tan θ or x = a cot θ |
| 3 | \(x^2 – a^2\) | x = a sec θ or x = a cosec θ |
| 4 | \(\sqrt{\frac{x+a}{a-x}}or\ \sqrt{\frac{a-x}{x+a}}\) | x = a cos 2θ |
| 5 | \(\sqrt{\frac{a^2+x^2}{a^2-x^2}}\ or\ \ \sqrt{\frac{a^2-x^2}{a^2+x^2}}\) | \(x^2 = a^2 cos 2θ\) |
| 6 | \(\frac{2x}{1+x^2}\ or\ \ \frac{2x}{1-x^2}\) | x = tan θ |
| 7 | a sin x + b cos x | a = r cos α, b = r sin α |
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Differentiation of Infinite Series
Consider if x is given in the form of an infinite series of y, then the subsequent steps are followed to find \(\frac{dy}{dx}\).
If \(x=\sqrt{f(y)+\sqrt{f(y)+\sqrt{f(y)+\dots.\infty}}}\)
Then \(x=\sqrt{f(y)+x}\)
\( \Rightarrow x^2=f(y)+x\)
\(\Rightarrow\ 2x\ \frac{dx}{dy}=f^{^{\prime}}\left(y\right)+\frac{dx}{dy}\)
\(\Rightarrow\ \ \frac{dx}{dy}=\frac{f^{^{\prime}}\left(y\right)}{2x-1}\)
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Differentiation on Determinants
Consider if:
\(\Delta\left(x\right)=\left|\begin{matrix}f_1\left(x\right)&g_1\left(x\right)\\
f_2\left(x\right)&g_2\left(x\right)\end{matrix}\right|\)
Then, to differentiate the above determinant, first, differentiate the first row/column keeping the other fixed, then apply an operation on the other row/column as explained below.
\(\Delta^{^{\prime}}\left(x\right)=\left|\begin{matrix}f_1^{^{\prime}}\left(x\right)&g_1^{^{\prime}}\left(x\right)\\
f_2\left(x\right)&g_2\left(x\right)\end{matrix}\right|+\left|\begin{matrix}f_1\left(x\right)&g_1\left(x\right)\\
f_2^{^{\prime}}\left(x\right)&g_2^{^{\prime}}\left(x\right)\end{matrix}\right|\)
OR
\(\Delta^{^{\prime}}\left(x\right)=\left|\begin{matrix}f_1^{^{\prime}}\left(x\right)&g_1\left(x\right)\\
f_2^{^{\prime}}\left(x\right)&g_2\left(x\right)\end{matrix}\right|+\left|\begin{matrix}f_1\left(x\right)&g_1^{^{\prime}}\left(x\right)\\
f_2\left(x\right)&g_2^{^{\prime}}\left(x\right)\end{matrix}\right|\)
