How do you find the equation for the tangent line for y=1x at x=1?

y=1x can be expressed in an alternate form as y=x−1 The derivative of y gives the slope of the tangent line at a point (x,y) dydx=(−1)x−2 or (equivalently) dydx=−1x2 At x=1 the slope is −112=−1 at x=1 y=1x=1 The tangent has a slope of (-1) and passes through the point (1,1) Using the slope-point form (y-1)=(-1)(x-1) y-1=1-x y=2-x The equation for the required tangent line is y=2-x

y=f(a)+f'(a)(x-a)where a is some arbitrary value and x is the typical unknown coordinate.If f(x)=1x:f(1)=11 and f′(1)=−1x2 at x=1⇒−112So:y=11−x−112ory=1-(x-1)=2-x is your tangent line.Power Rule: (f′(1x)=f′(x−1)=−1⋅x−2=−1x2)

y=1x can be expressed in an alternate form as y=x−1 The derivative of y gives the slope of the tangent line at a point (x,y) dydx=(−1)x−2 or (equivalently) dydx=−1x2 At x=1 the slope is −112=−1 at x=1 y=1x=1 The tangent has a slope of (-1) and passes through the point (1,1) Using the slope-point form (y-1)=(-1)(x-1) y-1=1-x y=2-x The equation for the required tangent line is y=2-x

y=f(a)+f'(a)(x-a)where a is some arbitrary value and x is the typical unknown coordinate.If f(x)=1x:f(1)=11 and f′(1)=−1x2 at x=1⇒−112So:y=11−x−112ory=1-(x-1)=2-x is your tangent line.Power Rule: (f′(1x)=f′(x−1)=−1⋅x−2=−1x2)

Step by step solution to "How do you find the equation for the..."

Frauffshiesiaf6s

Answered question

2022-02-13

How do you find the equation for the tangent line for

y = \frac{1}{x}

at x=1?

Answer & Explanation

mixtyggc

Beginner2022-02-14Added 15 answers

y = \frac{1}{x}

can be expressed in an alternate form as

y = x^{- 1}

The derivative of y gives the slope of the tangent line at a point (x,y)

\frac{d y}{d x} = (- 1) x^{- 2}

or (equivalently)

\frac{d y}{d x} = - \frac{1}{x^{2}}

At x=1
the slope is

\frac{- 1}{1^{2}} = - 1

at x=1

y = \frac{1}{x} = 1

The tangent has a slope of (-1) and passes through the point (1,1)
Using the slope-point form
(y-1)=(-1)(x-1)
y-1=1-x
y=2-x
The equation for the required tangent line is
y=2-x

Joel Molina

Beginner2022-02-15Added 9 answers

y=f(a)+f'(a)(x-a)
where a is some arbitrary value and x is the typical unknown coordinate.
If

f (x) = \frac{1}{x} :

f (1) = \frac{1}{1}

and

f^{'} (1) = - \frac{1}{x^{2}}

at x=1

\Rightarrow - \frac{1}{1^{2}}

So:

y = \frac{1}{1} - \frac{x - 1}{1^{2}}

or
y=1-(x-1)=2-x is your tangent line.
Power Rule:

(f^{'} (\frac{1}{x}) = f^{'} (x^{- 1}) = - 1 \cdot x^{- 2} = - \frac{1}{x^{2}})

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