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Question: Given that \(\dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right)\). The relationship \(f'\left( {a...

Given that ddxf(x)=f′(x)\dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right). The relationship f′(a+b)=f′(a)+f′(b)f'\left( {a + b} \right) = f'\left( a \right) + f'\left( b \right) is valid if f(x)f\left( x \right) is equal to

  1. xx
  2. x2{x^2}
  3. x3{x^3}
  4. x4{x^4}
Explanation

Solution

To solve this question, we will individually analyse each of the options and see which one of the following options satisfies the question. So, we will first differentiate the options and then substitute the differentiated values in the second condition, that is, f′(a+b)=f′(a)+f′(b)f'\left( {a + b} \right) = f'\left( a \right) + f'\left( b \right). The option that will satisfy this condition will be the correct answer.

Complete answer: So, the first condition is ddxf(x)=f′(x)\dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right).
And, f′(a+b)=f′(a)+f′(b)f'\left( {a + b} \right) = f'\left( a \right) + f'\left( b \right).
To find, which of the following options satisfies the second condition.
We will analyse the options individually to find the answer.
Analysing option 1:
f(x)=xf\left( x \right) = x
Differentiating both sides with respect to xx, we get,
f′(x)=1f'\left( x \right) = 1
So, here,
LHS: f′(a+b)=1f'\left( {a + b} \right) = 1
And, RHS: f′(a)+f′(b)=1+1=2f'\left( a \right) + f'\left( b \right) = 1 + 1 = 2
Therefore, LHS≠\neRHS.
Therefore, option 1 is not the correct option.
Analysing option 2:
f(x)=x2f\left( x \right) = {x^2}
Differentiating both sides with respect to xx, we get,
f′(x)=2xf'\left( x \right) = 2x
So, here,
LHS: f′(a+b)=2(a+b)f'\left( {a + b} \right) = 2\left( {a + b} \right)
And, RHS: f′(a)+f′(b)=2(a)+2(b)=2(a+b)f'\left( a \right) + f'\left( b \right) = 2\left( a \right) + 2\left( b \right) = 2\left( {a + b} \right)
Therefore, LHS==RHS.
Therefore, option 2 is the correct option.
Analysing option 3:
f(x)=x3f\left( x \right) = {x^3}
Differentiating both sides with respect to xx, we get,
f′(x)=3x2f'\left( x \right) = 3{x^2}
So, here,
LHS: f′(a+b)=3(a+b)2f'\left( {a + b} \right) = 3{\left( {a + b} \right)^2}
And, RHS: f′(a)+f′(b)=3a2+3b2f'\left( a \right) + f'\left( b \right) = 3{a^2} + 3{b^2}
Therefore, LHS≠\neRHS.
Therefore, option 3 is not the correct option.
Analysing option 4:
f(x)=x4f\left( x \right) = {x^4}
Differentiating both sides with respect to xx, we get,
f′(x)=4x3f'\left( x \right) = 4{x^3}
So, here,
LHS: f′(a+b)=4(a+b)3f'\left( {a + b} \right) = 4{\left( {a + b} \right)^3}
And, RHS: f′(a)+f′(b)=4a3+4b3f'\left( a \right) + f'\left( b \right) = 4{a^3} + 4{b^3}
Therefore, LHS≠\neRHS.
Therefore, option 4 is not the correct option.
So, the correct option is 2.

Note:
In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.