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)=√(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

`\frac{\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:

`\frac{\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^{(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.

`\frac{\mathrm{d}}{\mathrm{d}x}(\tan(x))=\sec^2(x)`

With this, we have that

`\frac{\mathrm{d}}{\mathrm{d}x}(w(x))=w^(\prime)(x)=(√(x))^(\prime)+(\tan(x))^(\prime)`

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

`w^(\prime)(x)=(1)/(2)\cdot\,x^{(1)/(2)-1}+\sec^2(x)=(1)/(2√(x))+\sec^2(x)`

This is the derivative of w.