Example 1: Equation of a Tangent Line

Find the equation of the tangent line to $f\left(x\right)=\sqrt{{5x}^4+{7x}^{–3}–10x}$ at x = 1.

Step 1: Find the tangent point, $\left(x_1,y_1\right)$.

You need to evaluate your given equation, $f\left(x\right)$, at the given x-value to find the y-value , $f\left(x_1\right)=y_1$.

In this example $x_1=1$. You need to find $f\left(1\right)$ in order to get our $y_1$.

$f\left(x\right)=\sqrt{{5x}^4+{7x}^{–3}–10x}$

$f\left(1\right)=\sqrt{{5\left(1\right)}^4+{7\left(1\right)}^{–3}–10\left(1\right)}$

$f\left(1\right)=\sqrt{2}$

$y_1=\sqrt{2}$

$\left(x_1,y_1\right)=(1,\sqrt{2})$

Step 2: Find the slope , m , of the tangent line.

The slope of the tangent line is found by evaluating your derivative at the given x-value , $f^{\prime }\left(x_1\right)=m=\mathit{slope}$.

To do this you will need to start by finding your actual $f\prime \left(x\right)$. Keep in mind that finding the derivative is now just a small part of a larger process.

Finding this derivative will require an algebraic rewrite (power over root) . After that you would follow the chain-rule process to get the derivative result, $f\prime \left(x\right)$.

Now to find the actual slope , m , of the tangent line, we will need to plug our x-value into that derivative equation, $f^{\prime }\left(1\right)=m=\mathit{slope}$.

$f\left(x\right)=\sqrt{{5x}^4+{7x}^{–3}–10x}$

$f\left(x\right)={({5x}^4+{7x}^{–3}–10x)}^{\frac{1}{2}}$

$f\prime \left(x\right)=({20x}^3–21x^{–4}–10)\frac{1}{2}{({5x}^4+{7x}^{–3}–10x)}^{–\frac{1}{2}}$

$f\prime \left(x\right)=({20x}^3–21x^{–4}–10)\frac{1}{2}{({5x}^4+{7x}^{–3}–10x)}^{–\frac{1}{2}}$

Calculus complete, begin algebra.

$f^{\prime }\left(1\right)=({20\left(1\right)}^3–21{\left(1\right)}^{–4}–10)\frac{1}{2}{({5\left(1\right)}^4+{7\left(1\right)}^{–3}–10(1\left)\right)}^{–\frac{1}{2}}$

$f^{\prime }\left(1\right)=(–11)\bullet \frac{1}{2}{\bullet \left(2\right)}^{–\frac{1}{2}}$

$f^{\prime }\left(1\right)=(–11)\bullet \frac{1}{2}\bullet \frac{1}{{\left(2\right)}^{\frac{1}{2}}}$

$f^{\prime }\left(1\right)=(–11)\bullet \frac{1}{2}\bullet \frac{1}{\sqrt{2}}$

$f^{\prime }\left(1\right)=\frac{–11}{2\sqrt{2}}$

$\mathit{slope}=m=\frac{–11}{2\sqrt{2}}$

Step 3: Find the equation of the tangent line.

You grab your point, $\left(x_1,y_1\right)=(1,\sqrt{2})$, from Step 1 , your $\mathit{slope}=m=\frac{–11}{2\sqrt{2}}$, from Step 2 , and you plug them into point-slope form of a line,
$y=m(x–x_1)+y_1$, to get the final result, the equation of the tangent line.

$\left(x_1,y_1\right)=(1,\sqrt{2})$

$\mathit{slope}=m=\frac{–11}{2\sqrt{2}}$

$y=m(x–x_1)+y_1$

$y=\frac{–11}{2\sqrt{2}}(x–1)+\sqrt{2}$

$y=\frac{–11}{2\sqrt{2}}(x–1)+\sqrt{2}$

Final Result:

The equation of the tangent line to the graph of $f\left(x\right)=\sqrt{{5x}^4+{7x}^{–3}–10x}$, at x = 1 , is $y=\frac{–11}{2\sqrt{2}}(x–1)+\sqrt{2}$.