Question

I found the derivative of the function using the quotient rule however I am lost from there

Answer 1

To answer this question we will use the quotient rule for derivatives:

$(\frac{h(x)}{g(x)})^{\prime}=\frac{h^{\prime}(x)g(x)-h(x)g^{\prime}(x)}{g(x)^2}\text{.}$

We know that:

$\begin{gathered} (\sqrt[]{x})^{\prime}=\frac{1}{2\sqrt[]{x}}, \\ (5x-6)^{\prime}=5. \end{gathered}$

Then:

$f^{\prime}(x)=\frac{\frac{1}{2\sqrt[]{x}}\cdot(5x-6)-\sqrt[]{x}\cdot5}{(5x-6)^2}\text{.}$

Therefore, the slope of the tangent line to the graph of f(x) at (1,f(1)) is:

$f^{\prime}(1)=\frac{\frac{1}{2\sqrt[]{1}}(5\cdot1-6)-\sqrt[]{1}\cdot5}{(5\cdot1-6)^2}\text{.}$

Simplifying the above result we get:

$\begin{gathered} f^{\prime}(1)=\frac{\frac{1}{2}(5-6)-5}{(5-6)^2} \\ =\frac{\frac{1}{2}(-1)-5}{(-1)^2}=\frac{-\frac{1}{2}-5}{1}=-\frac{11}{2}\text{.} \end{gathered}$

Now, we will use the following slope-point formula for the equation of a line:

$y-y_1=m(x-x_1)\text{.}$

Therefore the slope of the tangent line to the graph of f(x) at (1,f(1)) is:

$y-f(1)=-\frac{11}{2}(x-1)\text{.}$

Now, we know that:

$f(1)=\frac{\sqrt[]{1}}{5\cdot1-6}=\frac{1}{5-6}=\frac{1}{-1}=-1.$

Therefore:

$\begin{gathered} y-(-1)=-\frac{11}{2}(x-1), \\ y+1=-\frac{11}{2}x+\frac{11}{2}\text{.} \end{gathered}$

Subtracting 1 from the above equation we get:

$\begin{gathered} y+1-1=-\frac{11}{2}x+\frac{11}{2}-1, \\ y=-\frac{11}{2}x+\frac{9}{2}\text{.} \end{gathered}$

Answer:

$y=-\frac{11}{2}x+\frac{9}{2}\text{.}$