5.6 Practice: Finding a Tangent Line

Since we went through a lot of information in this topic, let’s put it all together to find the derivative of a function and then find a tangent line. We will do this for the following five equations:

1. $f(x)=2x^4-3x+1$

2. $f(x)=\sqrt{3x^2+1}$

3. $f(x)=e^x(2x^2-3x+2)$

4. $f(x)=\frac{x-2}{x^2+2x-4}$

5. $f(x)=\ln x^4-2x^3+3x^2-x$

For each function, find the derivative with respect to $x$, and then find the tangent line for each function that goes through $x=1$.

1. For the first function, we can find the derivative using the power rule:

$f^{\prime }(x)=8x^3-3x$

Therefore, the slope of the tangent line is

$m=f^{\prime }(x=1)=8\cdot 1^3-3\cdot 1=5$

When $x=1$, $y$ is

$f(x=1)=2\cdot 1^4-3\cdot 1+1=0$

This means that the tangent line goes through the point (1, 0). Using the point-slope formula, the equation of the tangent line is

$\begin{array}{rl}y& =5(x-1)\\ & =5x-5\end{array}$

2. For the second function, we need to use the chain rule to find the derivative:

$\begin{array}{rl}g(x)& =\sqrt{x}\\ h(x)& =3x^2+1\end{array}$

Therefore, the derivative of each function is

$\begin{array}{rl}{g}^{\prime }(x)& =\frac{1}{2\sqrt{x}}\\ h^{\prime }(x)& =6x\end{array}$

Using the chain rule, the derivative of $f(x)$ is

$\begin{array}{rl}{f}^{\prime }(x)& =h^{\prime }(x)g^{\prime }(h(x))\\ & =6x\cdot \frac{1}{2\sqrt{3x^2+1}}\\ & =\frac{3x}{\sqrt{3x^2+1}}\end{array}$

So the slope of the tangent line is

$f^{\prime }(x=1)=\frac{3\cdot 1}{\sqrt{3\cdot 1^2+1}}=\frac{3}{\sqrt{(}4)}=\frac{3}{2}$

We need to find the point of the tangent when $x=1$, so

$f(x=1)=\sqrt{3\cdot 1^2+1}=\sqrt{4}=2$

The tangent line has a slope of $\frac{3}{2}$  and goes through (1, 2). Thus, the equation of the tangent line is

$\begin{array}{rl}y-2& =\frac{3}{2}(x-1)\\ y& =\frac{3}{2}x+\frac{1}{2}\end{array}$

3. We need to use the product rule for the third function. Let $g$ and $h$ be

$\begin{array}{rl}g(x)& =e^x\\ h(x)& =2x^2-3x+2\end{array}$

Then the derivatives of $g$ and $h$ are

$\begin{array}{rl}{g}^{\prime }(x)& =e^x\\ h^{\prime }(x)& =4x-3\end{array}$

Now we apply the product rule to find the derivative of $f$:

$\begin{array}{rl}{f}^{\prime }(x)& =g^{\prime }(x)h(x)+g(x)h^{\prime }(x)\\ & =e^x(2x^2-3x+2)+e^x(4x-3)\\ & =e^x(2x^2+x-1)\end{array}$

Therefore, the slope of the line is

$m=f^{\prime }(x=1)=e^1(2\cdot 1^2+1-1)=2e$

The y-coordinate of the tangent point is

$f(x=1)=e^1(2\cdot 1^2-3\cdot 1+2)=e$

The tangent line has the slope of $\frac{2}{e}$  and goes through (1, $e$):

$\begin{array}{rl}y-e& =2e(x-1)\\ y& =2ex-e\end{array}$

4. The fourth function is a function divided by a function, so we need to use the quotient rule. We will set $g$ and $h$ to be

$\begin{array}{rl}g(x)& =x-2\\ h(x)& =x^2+2x-4\end{array}$

The derivatives of these two functions are

$\begin{array}{rl}{g}^{\prime }(x)& =1\\ h^{\prime }(x)& =2x+2\end{array}$

Now we can apply the quotient rule.

$\begin{array}{rl}{f}^{\prime }(x)& =\frac{g^{\prime }(x)h(x)-g(x)h^{\prime }(x)}{[h(x){]}^2}\\ & =\frac{1\cdot (x^2+2x-4)-(x-2)(2x+2)}{(x^2+2x-4{)}^2}\\ & =\frac{(x^2+2x-4)-(x-2)(2x+2)}{(x^2+2x-4{)}^2}\end{array}$

Even though we could simplify the derivative, we will leave it as it is since our purpose is to find the tangent line. The slope of the tangent line is

$\begin{array}{rl}m=f^{\prime }(x=1)& =\frac{(1^2+2\cdot 1-4)-(1-2)(2\cdot 1+2)}{(1^2+2\cdot 1-4{)}^2}\\ & =\frac{(-1)-(-1)(4)}{(-1{)}^2}\\ & =\frac{3}{1}\\ & =3\end{array}$

The y-coordinate of the tangent point is

$f(x=1)=\frac{1-2}{1^2+2\cdot 1-4}=\frac{-1}{-1}=1$

Therefore, the function of the tangent line is

$y-1=3(x-1)$
$y=3x-2$

5. Last but not least, we need to use the chain rule for the last function. We assign functions $g$ and $h$ as follows:

$g(x)=\ln x$

$h(x)=x^4-2x^3+3x^2-x$

The derivative of each function is

$g^{\prime }(x)=\frac{1}{x}$

$h^{\prime }(x)=4x^3-6x^2+6x-1$

Therefore, the derivative of the function $f$ is

$f^{\prime }(x)=h^{\prime }(x)g^{\prime }(h(x))$
$=\frac{4x^3-6x^2+6x-1}{x^4-2x^3+3x^2-x}$

The slope of the tangent line is

$m=f^{\prime }(x=1)=\frac{4\cdot 1^3-6\cdot 1^2+6\cdot 1-1}{1^4-2\cdot 1^3+3\cdot 1^2-1}$
$=\frac{3}{1}$
$=3$

Now the y-coordinate of the tangent point is

$f(x=1)=\ln (1^4-2\cdot 1^3+3\cdot 1^2-1)=\ln 1=0$

The equation of the tangent line with a slope of 3 that passes through the point (1, 0) is

$y=3(x-1)$
$=3x-3$