Respuesta :
Answer:
Differentiation of both the term is [tex]2x+3[/tex]
Step-by-step explanation:
As we have to use first principle of derivatives lets recall the formula.
[tex]f(x)= \lim_{h\to 0}\frac{(x+h)-f(x)}{h}[/tex]
Solving our eqaution.
[tex]y=x^2+3x[/tex]
We will work with [tex]f(x+h)[/tex] then [tex]-f(x)[/tex] separately then put in the above formula.
1.
[tex]f(x+h)=(x+h)^2+3(x+h)[/tex]
[tex](x^2+h^2+2hx+3x+3h)[/tex]
Now [tex]-f(x)[/tex]
[tex]-f(x)=-x^2-3x[/tex]
Plugging the values of both.
[tex]f(x)= \lim_{h\to 0}\frac{(x+h)-f(x)}{h}[/tex]
[tex]f(x)= \lim_{h\to 0}\frac{(x^2+h^2+2hx+3h+3x)- x^2-3x}{h}=\frac{h^2+2hx+3h}{h}[/tex]
Taking [tex]h[/tex] as common.
[tex]f(x)= \lim_{h\to 0}\frac{h^2+2hx+3h}{h}=h+2x+3[/tex]
Putting [tex]h=0[/tex]
Then
[tex]f(x)=2x+3[/tex] is the final derivative.
This will be same for [tex]y= x^2 + 3x + 8[/tex] as we have to put [tex]8[/tex] only.
2.
[tex]f(x+h)=(x+h)^2+3(x+h)+8[/tex]
[tex](x^2+h^2+2hx+3x+3h)[/tex]
Then [tex]-f(x)=-x^2-3x-8[/tex]
Plugging the values of both.
[tex]f(x)= \lim_{h\to 0}\frac{(x+h)-f(x)}{h}[/tex]
[tex]f(x)= \lim_{h\to 0}\frac{(x^2+h^2+2hx+3h+3x+8)- x^2-3x-8}{h}=\frac{h^2+2hx+3h}{h}[/tex]
Taking [tex]h[/tex] as common.
[tex]f(x)= \lim_{h\to 0}\frac{h^2+2hx+3h}{h}=h+2x+3[/tex]
Putting [tex]h=0[/tex]
Then
[tex]f(x)=2x+3[/tex] is the final derivative.
So both the derivatives are same.
