Question

Производная функция f(x)=x+6/COSx

Answer 1

$f(x) = \frac{x + 6}{ \cos(x) } \\ f \: '(x) = \frac{ \cos(x) \times (x + 6)'- ( \cos(x))'\times (x + 6) }{ \cos(x) {}^{2} } \:$

$f \: '(x) = \frac{ 1 \times \cos(x) - ( - \sin(x) \times (x + 6)) }{ \cos(x) {}^{2} } \:$

$f'(x) = \frac{ \cos(x) + 6 \sin(x) + \sin(x) \times x }{ { \cos(x) }^{2} }$

Answer 2

To find the derivative of the function $f(x) = x + \frac{6}{\cos(x)}$, we can use the rules of differentiation. The derivative of a sum is the sum of the derivatives, and we can use the quotient rule to find the derivative of $\frac{6}{\cos(x)}$. Let's go step by step:

Step 1: Find the derivative of $x$ with respect to $x$. $\frac{d}{dx}(x) = 1.$

Step 2: Find the derivative of $\frac{6}{\cos(x)}$ with respect to $x$ using the quotient rule.

The quotient rule states that for a function $\frac{u(x)}{v(x)}$, the derivative is given by: $\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x)\frac{du(x)}{dx} - u(x)\frac{dv(x)}{dx}}{(v(x))^2}.$

For our function $\frac{6}{\cos(x)}$, let $u(x) = 6$ and $v(x) = \cos(x)$.

The derivatives of $u(x)$ and $v(x)$ are: $\frac{du(x)}{dx} = 0 \quad \text{(the derivative of a constant is 0)},$ $\frac{dv(x)}{dx} = -\sin(x) \quad \text{(derivative of }\cos(x)\text{ is }-\sin(x)).$

Now, applying the quotient rule: $\frac{d}{dx}\left(\frac{6}{\cos(x)}\right) = \frac{\cos(x)(0) - 6(-\sin(x))}{\cos^2(x)} = \frac{6\sin(x)}{\cos^2(x)}.$

So, the derivative of $f(x) = x + \frac{6}{\cos(x)}$ with respect to $x$ is: $f'(x) = 1 + \frac{6\sin(x)}{\cos^2(x)}.$

You can simplify this further if needed, but that's the derivative of the given function.