Respuesta :
Answer:
The derivative of the function f(x) is:
[tex]f'(x)=-2x[/tex]
Step-by-step explanation:
We are given a function f(x) as:
[tex]f(x)=5-x^2[/tex]
We have:
[tex]f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)[/tex]
( Since,
[tex](a+b)^2=a^2+b^2+2ab[/tex] )
Hence, we get:
[tex]f(x+h)=5-x^2-h^2-2xh[/tex]
Also, by using the definition of f'(x) i.e.
[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}[/tex]
Hence, on putting the value in the formula:
[tex]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[/tex]
Hence, the derivative of the function f(x) is:
[tex]f'(x)=-2x[/tex]
Answer:
The derivative of given function is -2x.
Step-by-step explanation:
The first principle of differentiation is
[tex]f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]
The given function is
[tex]f(x)=5-x^2[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-(x+h)^2-(5-h^2}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-(x^2+2xh+h^2)-5+h^2}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-x^2-2xh-h^2-5+h^2}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2-2xh}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2}{h}-\frac{2xh}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2}{h}-2x[/tex]
Apply limit.
[tex]f'(x)=\frac{-x^2}{0}-2x[/tex]
[tex]f'(x)=0-2x[/tex]
[tex]f'(x)=-2x[/tex]
Therefore, the derivative of given function is -2x.
