Question

Найти производную функции y= -150 In cos x

Answer 1

Ответ:

150tgx

Пошаговое объяснение:

у=-150(ln cosx)

y' = -150(ln cosx)' = -150(1/cosx)*(cosx)' = -(-150sinx)/(cosx) = 150tgx

Answer 2

To find the derivative of the function $y = -150 \ln(\cos(x))$, we will use the chain rule and the derivative of $\ln(u)$, which is $\frac{{du}}{{dx}}$.

Let's break it down step-by-step:

Step 1: Identify the inner function ($u$). The inner function is $\cos(x)$.

Step 2: Find the derivative of the inner function ($\frac{{du}}{{dx}}$). The derivative of $\cos(x)$ is $-\sin(x)$.

Step 3: Apply the chain rule. The chain rule states that if $y = f(u)$ and $u = g(x)$, then $\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}$.

In our case: $f(u) = -150\ln(u)$ and $u = \cos(x)$.

So, $\frac{{dy}}{{du}} = -150 \cdot \frac{{d}}{{du}}(\ln(u)) = -150 \cdot \frac{1}{u} = -\frac{{150}}{{u}}$.

Step 4: Substitute the values back into the chain rule formula. $\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} = -\frac{{150}}{{\cos(x)}} \cdot (-\sin(x))$.

Step 5: Simplify the expression. $\frac{{dy}}{{dx}} = \frac{{150\sin(x)}}{{\cos(x)}}$.

Now, we have the derivative of the function $y = -150 \ln(\cos(x))$: $y' = \frac{{150\sin(x)}}{{\cos(x)}}$.

Alternatively, we can rewrite the derivative using the identity $\tan(x) = \frac{{\sin(x)}}{{\cos(x)}}$:

$y' = 150\cdot \tan(x)$.