## problem statement

Find the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.

$y=3 \sqrt[4]{x}-\sqrt{x}, x_{0}=1$
## Solution

Let's find $y^{\prime}$:

$y^{\prime}=(3 \sqrt[4]{x}-\sqrt{x})^{\prime}=\left(3 x^{\frac{1}{4}}-\sqrt{x}\right)^{\prime}=3 \cdot \frac{1}{4} \cdot x^{-\frac{3}{4}}-\frac{1}{2 \sqrt{x}}=\frac{3}{4 \sqrt[4]{x^{3}}}-\frac{1}{2 \sqrt{x}}$

Then

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{3}{4 \sqrt[4]{1^{3}}}-\frac{1}{2 \sqrt{1}}=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}$

Since the function $y^{\prime}$ at the point $x_{0}$ has a finite derivative, the equation of the tangent line is:

$y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right)$, where

$y_{0}^{\prime}=\frac{1}{4}$

$y_{0}=y\left(x_{0}\right)=3 \sqrt[4]{1}-\sqrt{1}=3-1=2$

We get:

$y-2=\frac{1}{4} \cdot(x-1)$

$y=\frac{1}{4} \cdot x-\frac{1}{4}+2$

$y=\frac{x}{4}+\frac{7}{4}$

Thus, the equation of the tangent line is:

$y=\frac{x}{4}+\frac{7}{4}$

## Problem Kuznetsov Differentiation 3-27