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

How do I use the limit definition of derivative to find f′(x)f'(x) for f(x)=5x−9x2f(x) = 5x - 9{x^2} ?

Explanation

Solution

Hint : Here we need to find the derivative of the given function using the limit definition. We have f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} . Here we have f(x)=5x−9x2f(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−9x2f(x) = 5x - 9{x^2} .
From the definition of the derivatives we have,
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} .
We need f(x+h)f(x + h) , now we find f(x+h)f(x + h) .
f(x+h)=5(x+h)−9(x+h)2f(x + h) = 5(x + h) - 9{(x + h)^2}
Now using (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab we have,
=5(x+h)−9(x2+h2+2xh)= 5(x + h) - 9({x^2} + {h^2} + 2xh)
Now expanding the brackets we have,
=5x+5h−9x2−9h2−18xh= 5x + 5h - 9{x^2} - 9{h^2} - 18xh .
Thus we have,
f(x+h)=5x+5h−9x2−9h2−18xhf(x + h) = 5x + 5h - 9{x^2} - 9{h^2} - 18xh .
Now substituting we have,
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}
f′(x)=lim⁡h→0(5x+5h−9x2−9h2−18xh)−(5x−9x2)hf'(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,
=lim⁡h→05x+5h−9x2−9h2−18xh−5x+9x2h= \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,
=lim⁡h→05h−9h2−18xhh= \mathop {\lim }\limits_{h \to 0} \dfrac{{5h - 9{h^2} - 18xh}}{h}
Now taking h common in the numerator we have,
=lim⁡h→0h(5−9h−18x)h= \mathop {\lim }\limits_{h \to 0} \dfrac{{h\left( {5 - 9h - 18x} \right)}}{h}
=lim⁡h→0(5−9h−18x)= \mathop {\lim }\limits_{h \to 0} \left( {5 - 9h - 18x} \right)
Now applying the limit as ‘h’ tends to zero we have,
=(5−9(0)−18x)= \left( {5 - 9(0) - 18x} \right)
=5−18x= 5 - 18x
Thus we have ⇒f′(x)=5−18x \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 xn{x^n} with respect to ‘x’ is d(xn)dx=nxn−1\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}} .
We have, f(x)=5x−9x2f(x) = 5x - 9{x^2} . Differentiation with respect to ‘x’ and using the formula we have,
f′(x)=5(1×x1−1)−9(2×x2−1)f'(x) = 5\left( {1 \times {x^{1 - 1}}} \right) - 9\left( {2 \times {x^{2 - 1}}} \right)
f′(x)=5(1×x0)−9(2×x1)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.
⇒f′(x)=5−18x\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.