Question

Identify each expression that represents the slope of a tangent to the curve y=1/x+1 at any point (x,y) .

Answer 1

Answer:

Slope of a tangent to the curve = $f'(x) = \frac{-1 }{(x+1)^{2} }$

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

We have,

f(x) = y = $\frac{1}{x+1}$

So,

f(x+h) = $\frac{1}{x+h+1}$

Now,

Slope = f'(x)

And

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\= \lim_{h \to 0} \frac{\frac{1}{x+h+1} - \frac{1}{x+1} }{h}\\= \lim_{h \to 0} \frac{x+1 - (x+h+1) }{h(x+1)(x+h+1)}\\= \lim_{h \to 0} \frac{x +1 - x-h-1 }{h(x+1)(x+h+1)}\\= \lim_{h \to 0} \frac{-h }{h(x+1)(x+h+1)}\\= \lim_{h \to 0} \frac{-1 }{(x+1)(x+h+1)}\\= \frac{-1 }{(x+1)(x+0+1)}\\= \frac{-1 }{(x+1)(x+1)}\\= \frac{-1 }{(x+1)^{2} }$

∴ we get

Slope of a tangent to the curve = $f'(x) = \frac{-1 }{(x+1)^{2} }$