How do you find the equation of the tangent line to the curve y=2x+1x+2 at the given point (1,1)?

We have, y=2x+1x+2 Now, the derivative (using quotient rule) dydx=2⋅(x+2)−(2x+1)(x+2)2=3(x+2)2 Now dydx at (1,1) is the slope of the tangent to the curve at (1,1) m=(dydx)1,1=39=13 Now, point-slope form a straight line is given by (y−y1)=m⋅(x−x1) Now, (x1,y1)=(1,1) and m=13, substituting these values in above equation we get, y−1=13⋅(x−1)⇒y=x3+23 (Answer)

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High School
Calculus and Analysis
Calculus 1
Derivatives

Cassidy Johnson

Answered question

2022-02-12

How do you find the equation of the tangent line to the curve

y = \frac{2 x + 1}{x + 2}

at the given point (1,1)?

Answer & Explanation

Rachel Frazier

Beginner2022-02-13Added 14 answers

We have,

y = \frac{2 x + 1}{x + 2}

Now, the derivative (using quotient rule)

\frac{d y}{d x} = \frac{2 \cdot (x + 2) - (2 x + 1)}{{(x + 2)}^{2}} = \frac{3}{{(x + 2)}^{2}}

Now

\frac{d y}{d x}

at (1,1) is the slope of the tangent to the curve at (1,1)

m = {(\frac{d y}{d x})}_{1, 1} = \frac{3}{9} = \frac{1}{3}

Now, point-slope form a straight line is given by

(y - y_{1}) = m \cdot (x - x_{1})

Now,

(x_{1}, y_{1}) = (1, 1)

and

m = \frac{1}{3}

, substituting these values in above equation we get,

y - 1 = \frac{1}{3} \cdot (x - 1) \Rightarrow y = \frac{x}{3} + \frac{2}{3}

(Answer)

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