Question

How do I use the limit definition of derivative to find \[f'(x)\] for \[f(x) = 5x - 9{x^2}\] ?

Answer 1

Hint: Here we need to find the derivative of the given function using the limit definition. We have \[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\] . Here we have \[f(x) = 5x - 9{x^2}\] by substituting this In the formula we will get the desired answer.

Complete step by step solution:
Given,
 \[f(x) = 5x - 9{x^2}\] .
From the definition of the derivatives we have,
 \[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\] .
We need \[f(x + h)\] , now we find \[f(x + h)\] .
 \[f(x + h) = 5(x + h) - 9{(x + h)^2}\]
Now using \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] we have,
 \[ = 5(x + h) - 9({x^2} + {h^2} + 2xh)\]
Now expanding the brackets we have,
 \[ = 5x + 5h - 9{x^2} - 9{h^2} - 18xh\] .
Thus we have,
 \[f(x + h) = 5x + 5h - 9{x^2} - 9{h^2} - 18xh\] .
Now substituting we have,
 \[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\]
 \[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {5x + 5h - 9{x^2} - 9{h^2} - 18xh} \right) - \left( {5x - 9{x^2}} \right)}}{h}\] .
If we apply the limit we will get indeterminate form so we simply this further,
 \[ = \mathop {\lim }\limits_{h \to 0} \dfrac{{5x + 5h - 9{x^2} - 9{h^2} - 18xh - 5x + 9{x^2}}}{h}\]
Cancelling terms we have,
 \[ = \mathop {\lim }\limits_{h \to 0} \dfrac{{5h - 9{h^2} - 18xh}}{h}\]
Now taking h common in the numerator we have,
 \[ = \mathop {\lim }\limits_{h \to 0} \dfrac{{h\left( {5 - 9h - 18x} \right)}}{h}\]
 \[ = \mathop {\lim }\limits_{h \to 0} \left( {5 - 9h - 18x} \right)\]
Now applying the limit as ‘h’ tends to zero we have,
 \[ = \left( {5 - 9(0) - 18x} \right)\]
 \[ = 5 - 18x\]
Thus we have \[ \Rightarrow f'(x) = 5 - 18x\]
So, the correct answer is “ 5 - 18x”.

Note: We can solve this if we know the differentiation formula. That is the differentiation of \[{x^n}\] with respect to ‘x’ is \[\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}\] .
We have, \[f(x) = 5x - 9{x^2}\] . Differentiation with respect to ‘x’ and using the formula we have,
 \[f'(x) = 5\left( {1 \times {x^{1 - 1}}} \right) - 9\left( {2 \times {x^{2 - 1}}} \right)\]
 \[f'(x) = 5\left( {1 \times {x^0}} \right) - 9\left( {2 \times {x^1}} \right)\]
We know that zero power of any number is one.
 \[ \Rightarrow f'(x) = 5 - 18x\] . Thus in both the cases we have the same answer. But they particularly asked us to solve using the limit definition so we need to solve the above one.