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Disguised Derivatives – Advanced differentiation | Class 12 Maths
  • Last Updated : 02 Feb, 2021

The dictionary meaning of “disguise” is “unrecognizable”. Disguised derivative means “unrecognized derivative”. In this type of problems, the definition of derivative is hidden in the form of limit. At a glance, the problem seems to be solvable using limit properties but it is much easier to solve it using the first principle of derivative. So, this type of problems are known as derivative in disguise.

Before proceeding further, we need to revise the first principle of derivative. Finding the derivative of a function by computing this limit is known as differentiation from first principles. Derivative by first principle refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change, which is equal to 

\begin{aligned} \ \ \ \ \ \ \ \ \ f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \end{aligned}

Method to solve 

Step 1.Try to simplify the equation in such a way that it resemble the first principle of derivative.

Step 2.Find the value of f(x)



Step 3.Find f'(x) using first principle.

Step 4.Find the value of x such that on substituting it resembles the question

Let’s understand it better using an example problem.

Sample Problems On Disguised Derivatives

Example 1. \begin{aligned}  \qquad &\lim _{h \rightarrow 0} \frac{5(2+h)^{3}-40}{h} \end{aligned}

Solution:

This type of problems can be solved using limit evaluation as well as differentiation. Let us understand both solution one by one and later compare which one is better.


Solution using limit evaluation:

 \\ \begin{array}{l} \qquad =\lim _{h \rightarrow 0} \frac{5\left(2^{3}+h^{3}+3.2 \cdot h(2+h)\right)-40}{h} \\ \qquad =\lim _{h \rightarrow 0} \frac{40+5 h^{3}+30 h(2+h)-40}{h} \\ \qquad =\lim _{h \rightarrow 0} \frac{5 h^{3}+60 h+30 h^{2}}{h} \\ \qquad =\lim _{h \rightarrow 0} 5 h^{2}+60+30 h \\ \qquad =0+60+0 \\ \qquad =60 \end{array}



Now we have the answer to our problem i.e. 60. This function was not too complicated and limit evaluation was not that difficult. So, this approach works pretty fine. But in harder variations it would be really complicated to evaluate the limit. Our main intention in the second method would be to try to simplify the equation in such a way that it resembles the first principle of derivative.

Solution using differentiation:

Simplify the given problem:

\\ \qquad=\lim _{h \rightarrow 0} \frac{5(2+h)^{3}-40}{h} \\ \qquad=5 \cdot \lim _{h \rightarrow 0} \frac{(2+h)^{3}-8}{h} \\ \qquad=5 \cdot \lim _{h \rightarrow 0} \frac{(2+h)^{3}-2^{3}}{h} \\ \qquad=\text { Ans }
Let’s take f(x)=x3
\\ f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \quad\{first principle \} \\f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{(x+h)^{3}-x^{3}}{h} \quad\left\{f(x+h)=(x+h)^{3}\right\}
Putting x=2 on both sides,
\\f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{(2+h)^{3}-2^{3}}{h}
Multiply both sides by 5,
 \\5 . f^{\prime}(2)=5 . \lim _{h \rightarrow 0} \frac{(2+h)^{3}-2^{3}}{h} \\5 . f^{\prime}(2) =A n s \\ \ \ \ \ \ A n s =5 . f^{\prime}(2) \\ \ \ \ \ \ \ \ \ \ \   =5 .\left(3 x^{2}\right)_{a t} x=2 \\ \ \ \ \ \ \ \ \ \ \   =5 .(3.2 .2) \\ \ \ \ \ \  \ \ \ \ \  =60

The beauty of this method is all about knowing f(x). The complicated looking limit was a derivative in disguise. So, if you have any difficulty understanding the process go through it again. It would be much clearer after solving a few more examples.

Example 2. \\ \qquad\lim _{h \rightarrow 0} \frac{2 \tan \left(\frac{\pi}{3}+h\right)-2 \tan \left(\frac{\pi}{3}\right)}{h}

Solution:

Take f(x) = tan(x)
\\ \qquad f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\qquad \quad\{\text { first principle }\} \\ \ \ \qquad\qquad=\lim _{h \rightarrow 0} \frac{\tan (x+h)-\tan (x)}{h}
Put x=\frac{\pi}{3} and multiply both sides by 2,
\\ \qquad2. f^{\prime}\left(\frac{\pi}{3}\right)=\lim _{h \rightarrow 0} \frac{2 \tan \left(\frac{\pi}{3}+h\right)-2 \tan \left(\frac{\pi}{3}\right)}{h}\\ \qquad \qquad A n s =2 . f^{\prime}\left(\frac{\pi}{3}\right) \\ \qquad \qquad \qquad =2 . \sec ^{2}\left(\frac{\pi}{3}\right) \\ \qquad \qquad \qquad =2.2^{2} \\ \qquad \qquad \qquad =8

Example 3. \\ \begin{aligned} \qquad &\lim _{h \rightarrow 0} \frac{(x+h)^{2}-x^2}{h} \end{aligned}

Solution:

Let’s take f(x) = x2
\\ f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \quad\{first principle \} \\f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{(x+h)^{2}-x^{2}}{h} \quad\left\{f(x+h)=(x+h)^{2}\right\} \\Hence, \\ \qquad Ans=f'(x) \\ \qquad \qquad \ =2x

Example 4.\\ \lim _{h \rightarrow 0} \frac{ \sin \left(\frac{\pi}{3}+h\right)- \sin \left(\frac{\pi}{3}\right)}{h}

Solution:

Take f(x) = sin(x)
\\ \qquad f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\qquad \quad\{\text { first principle }\} \\ \ \ \qquad\qquad=\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin (x)}{h} \\ \qquad Put\ x=\frac{\pi}{3}\ \\ \qquad f^{\prime}\left(\frac{\pi}{3}\right)=\lim _{h \rightarrow 0} \frac{ \sin \left(\frac{\pi}{3}+h\right)- \sin \left(\frac{\pi}{3}\right)}{h} \\ \qquad \qquad A n s = f^{\prime}\left(\frac{\pi}{3}\right) \\ \qquad \qquad \qquad = \cos\left(\frac{\pi}{3}\right) \\ \qquad \qquad \qquad =1/2

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