Question

I need help question

Answer 1

Hello there. To solve this question, we have to remember some properties about algebraic structure of a function and taking derivatives.

Given the function:

$w(x)=\sqrt{x}+\tan(x)$

As you can see, it is composed by the addition of two functions, with different properties. One is the square root of a variable, while the other is the tangent of this variable, a trigonometric function.

In this case, we say that

$w(x)\text{ is a sum }f(x)+g(x)\text{ of basic functions}$

Now, to compute its derivative, remember that

The differential operator is linear, which means that the derivative of a sum of two functions is equivalent to the sum of the derivatives of the functions

$\dfrac{\mathrm{d}}{\mathrm{d}x}(f(x)+g(x))=f^{\prime}(x)+g^{\prime}(x)$

The derivative of a power is calculated by the power rule, given as follows:

$\dfrac{\mathrm{d}}{\mathrm{d}x}(x^n)=n\cdot\,x^{n-1},\text{ where }n\in\mathbb{R}$

In this case, remember that

$\begin{equation*} \large{\sqrt[n]{x^m}=x^{\frac{m}{n}}} \end{equation*}$

The derivative of the tangent function is equal to secant squared, that can be shown using the quotient rule and the derivative of sine and cosine functions.

$\dfrac{\mathrm{d}}{\mathrm{d}x}(\tan(x))=\sec^2(x)$

With this, we have that

$\dfrac{\mathrm{d}}{\mathrm{d}x}(w(x))=w^{\prime}(x)=(\sqrt{x})^{\prime}+(\tan(x))^{\prime}$

Applying the power rule and taking the derivative of the tangent, we get

$w^{\prime}(x)=\dfrac{1}{2}\cdot\,x^{\frac{1}{2}-1}+\sec^2(x)=\dfrac{1}{2\sqrt{x}}+\sec^2(x)$

This is the derivative of w.