Answer :
Answer:
Slope of a tangent to the curve = [tex]f'(x) = \frac{-1 }{(x+1)^{2} }[/tex]
Step-by-step explanation:
Given - y = 1/x+1
To find - Identify each expression that represents the slope of a tangent to the curve y=1/x+1 at any point (x,y) .
Proof -
We know that,
Slope of tangent line = f'(x) = [tex]\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}[/tex]
We have,
f(x) = y = [tex]\frac{1}{x+1}[/tex]
So,
f(x+h) = [tex]\frac{1}{x+h+1}[/tex]
Now,
Slope = f'(x)
And
[tex]f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\= \lim_{h \to 0} \frac{\frac{1}{x+h+1} - \frac{1}{x+1} }{h}\\= \lim_{h \to 0} \frac{x+1 - (x+h+1) }{h(x+1)(x+h+1)}\\= \lim_{h \to 0} \frac{x +1 - x-h-1 }{h(x+1)(x+h+1)}\\= \lim_{h \to 0} \frac{-h }{h(x+1)(x+h+1)}\\= \lim_{h \to 0} \frac{-1 }{(x+1)(x+h+1)}\\= \frac{-1 }{(x+1)(x+0+1)}\\= \frac{-1 }{(x+1)(x+1)}\\= \frac{-1 }{(x+1)^{2} }[/tex]
∴ we get
Slope of a tangent to the curve = [tex]f'(x) = \frac{-1 }{(x+1)^{2} }[/tex]
