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:

`((h(x))/(g(x)))^(\prime)=(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)=((1)/(2)(5-6)-5)/((5-6)^2) \\ =((1)/(2)(-1)-5)/((-1)^2)=(-(1)/(2)-5)/(1)=-(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)=-(11)/(2)(x-1)\text{.}`

Now, we know that:

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

Therefore:

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

Subtracting 1 from the above equation we get:

`\begin{gathered} y+1-1=-(11)/(2)x+(11)/(2)-1, \\ y=-(11)/(2)x+(9)/(2)\text{.} \end{gathered}`

Answer:

`y=-(11)/(2)x+(9)/(2)\text{.}`