What is the equation of the tangent line of f(x)=x+15−xx−15 at x=5?

Explanation:Tangent Line to a Curvey=x+15−xx−15Substitute for x value to get the coordinates of the point which lies on the tangenty=1+452Differentiate the function UsingPower rule ,Quotient rule , Chain ruley′=12x+15−(x−15)×1−x×1(x−15)2Simplifyy′=12x+15+15(x−15)2Now We substitute for x=5 to get the Slope of the Tangent myx=5′=3+520=mnow Substitute in the straight line equationy−y1=m(x−x1)y−1+452=3+520(x−5)20y−(3+5)x+5−355=0

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jncuenodd4nf

Answered question

2022-03-25

What is the equation of the tangent line of

f (x) = \sqrt{x + 15} - \frac{x}{x - 15}

at x=5?

Answer & Explanation

arrebusnaavbz

Beginner2022-03-26Added 18 answers

Explanation:
Tangent Line to a Curve

y = \sqrt{x + 15} - \frac{x}{x - 15}

Substitute for x value to get the coordinates of the point which lies on the tangent

y = \frac{1 + 4 \sqrt{5}}{2}

Differentiate the function Using
Power rule ,Quotient rule , Chain rule

y^{'} = \frac{1}{2 \sqrt{x + 15}} - \frac{(x - 15) \times 1 - x \times 1}{{(x - 15)}^{2}}

Simplify

y^{'} = \frac{1}{2 \sqrt{x + 15}} + \frac{15}{{(x - 15)}^{2}}

Now We substitute for x=5 to get the Slope of the Tangent m

y_{x = 5}^{'} = \frac{3 + \sqrt{5}}{20} = m

now Substitute in the straight line equation

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

y - \frac{1 + 4 \sqrt{5}}{2} = \frac{3 + \sqrt{5}}{20} (x - 5)

20 y - (3 + \sqrt{5}) x + 5 - 35 \sqrt{5} = 0

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