Question

How do you find $f'\left( 2 \right)$ using the limit definition given $f\left( x \right)=9-{{x}^{2}}$?

Answer 1

The derivative of any function $f\left( x \right)$ at point c can be given by the value of $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( c+h \right)-f\left( c \right)}{h}$ . Hence in the given problem we will substitute c = 2 and use the function to write the definition of derivative with limits. Then we will simplify the expression and hence find the value of the derivative.

Complete step by step solution:
Now consider the given function. $f\left( x \right)=9-{{x}^{2}}$ .
We want to find the derivative of the function at x = 2. Derivative of a function at a point is nothing but slope of the function at that point.
The derivative of a function at point c is given by $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( c+h \right)-f\left( c \right)}{h}$ . Now we will use this definition to find the derivative of the function at 2. Hence we have,
$\Rightarrow f'\left( 2 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 2+h \right)-f\left( 2 \right)}{h}$
Now substituting the values in the function given which is $f\left( x \right)=9-{{x}^{2}}$ we get,
$\Rightarrow f'\left( 2 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{9-{{\left( 2+h \right)}^{2}}-\left( 9-{{2}^{2}} \right)}{h}$
Now we know that the expansion of ${{\left( a+b \right)}^{2}}$ is given by ${{a}^{2}}+2ab+{{b}^{2}}$ . Hence using this we get,
$\Rightarrow f'\left( 2 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{9-\left( 4+4h+{{h}^{2}} \right)-\left( 9-4 \right)}{h}$
Opening the brackets in the equation we get,

& \Rightarrow f'\left( 2 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{9-4-4h-{{h}^{2}}-5}{h} \\\ & \Rightarrow f'\left( 2 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-4h-{{h}^{2}}}{h} \\\ \end{aligned}$$ Now taking h common from the numerator we get, $$\Rightarrow f'\left( 2 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-h\left( 4+h \right)}{h}$$ Cancelling h from numerator and denominator we get, $$\Rightarrow f'\left( 2 \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( -4-h \right)=-4$$ **Hence we get the value of $f'\left( 2 \right)=-4$.** **Note:** Now note that we can check the solution by using the derivative formulas. We know that the derivative of $9-{{x}^{2}}$ is given by $0-2x$ as derivative of constant is 0 and derivative of ${{x}^{n}}$ is given by $n{{x}^{n-1}}$ . Hence substituting x = 2 we get $f'\left( 2 \right)=-4$ which is the same as the solution obtained.