Question

Finding an Equation of a Tangent Line In Exercise, find an equation of the tangent line to the graph of the function at the given point. See Example 5.
f(x) = ln{x(x + 3)^1/2}; (6/5, 9/10)

Answer 1

Answer:

$75x-70y=27$

Step-by-step explanation:

We are given that

$y=ln(x(x+3)^{\frac{1}{2}})$

Point ($\frac{6}{5},\frac{9}{10}$)

We have to find the equation of tangent line to the given graph.

Differentiate w.r.t x

$\frac{dy}{dx}=\frac{1}{x(x+3)^{\frac{1}{2}}}\times ((x+3)^{\frac{1}{2}+x\times \frac{1}{2}(x+3)^{-\frac{1}{2}})$

By using formula

$\frac{d(lnx)}{dx}=\frac{1}{x}$

$\frac{dx^n}{dx}=nx^{n-1}$

$\frac{d(u\cdot v)}{dx}=u'v+v'u$

$\frac{dy}{dx}=\frac{x+3+x}{x(x+3)}=\frac{2x+3}{x(x+3)}$

Substitute x=6/5

$m=\frac{dy}{dx}=\frac{\frac{12}{5}+3}{\frac{6}{5}(\frac{6}{5}+3)}=\frac{15}{14}$

$m=\frac{dy}{dx}=\frac{15}{14}$

Slope-point form:

$y-y_1=m(x-x_1)$

$x_1=6/5,y_1=9/10$

By using this formula

The equation of tangent to the given graph

$y-\frac{9}{10}=\frac{15}{14}(x-\frac{6}{5})$

$\frac{10y-9}{10}=\frac{15}{14}(x-\frac{6}{5})$

$140y-126=150x-180$

$150x-140y=180-126=54$

$75x-70y=27$

The equation of tangent to the given graph

$75x-70y=27$