Question

F(x)=x-tg(-2x)= Плизз помогите с производными

Answer 1

$f(x) =x - \tan( - 2x) \\ f(0) = 0 - \tan( - 2 \times 0) \\ f(0) = - \tan(0 ) \\ f(0) = - 0 \\ f(0) = 0$

Answer 2

Sure thing! Let's work through it. The function $f(x) = x - \tan(-2x)$ involves both a linear term $x$ and a trigonometric term $\tan(-2x)$. To find the derivative, we'll use the sum/difference rule and the chain rule.

1. Derivative of the linear term $x$: $\frac{d}{dx}(x) = 1$

2. Derivative of the trigonometric term $\tan(-2x)$: Let $u = -2x$, then $\frac{du}{dx} = -2$. $\frac{d}{dx}(\tan(u)) = \sec^2(u) \cdot \frac{du}{dx}$ $= \sec^2(-2x) \cdot (-2)$

Now, applying the sum/difference rule, the derivative $f'(x)$ is the sum of the derivatives of the individual terms: $f'(x) = 1 - \sec^2(-2x) \cdot 2$

Keep in mind that the secant function is the reciprocal of the cosine function: $\sec(x) = \frac{1}{\cos(x)}$. So, you can rewrite the expression in terms of cosine.

If you have any specific values of $x$ where you want to evaluate the derivative or if you have any further questions, feel free to ask!