Question

Use f’( x ) = lim With h ---> 0 [f( x + h ) - f ( x )]/h to find the derivative at x for the given function. 5-x²

Answer 1

Answer:

The derivative of the function f(x) is:

                 $f'(x)=-2x$

Step-by-step explanation:

We are given a function f(x) as:

$f(x)=5-x^2$

We have:

$f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)$

( Since,

$(a+b)^2=a^2+b^2+2ab$ )

Hence, we get:

$f(x+h)=5-x^2-h^2-2xh$

Also, by using the definition of f'(x) i.e.

$f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

Hence, on putting the value in the formula:

$f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x$

      Hence, the derivative of the function f(x) is:

          $f'(x)=-2x$

Answer 2

Answer:

The derivative of given function is -2x.

Step-by-step explanation:

The first principle of differentiation is

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

The given function is

$f(x)=5-x^2$

$f'(x)=lim_{h\rightarrow 0}\frac{5-(x+h)^2-(5-h^2}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{5-(x^2+2xh+h^2)-5+h^2}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{5-x^2-2xh-h^2-5+h^2}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{-x^2-2xh}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{-x^2}{h}-\frac{2xh}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{-x^2}{h}-2x$

Apply limit.

$f'(x)=\frac{-x^2}{0}-2x$

$f'(x)=0-2x$

$f'(x)=-2x$

Therefore, the derivative of given function is -2x.