Question

Find the derivative of y=(tsint)/(1+t)

Answer 1

Task: Find the derivative of

$$y = \frac {\mathrm {t} \cdot \sin t}{1 + t}$$

Solution: The derivative of $y$ is $y'(t)$ :

$$y ^ {\prime} (t) = \frac {d}{d t} \left(\frac {\mathrm {t} \cdot \sin t}{1 + t}\right)$$

Use the quotient rule, $\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dt} - u\frac{dv}{dt}}{v^2}$ , where $u = t \cdot \sin t$ and $v = t + 1$ :

$$y ^ {\prime} (t) = \frac {(1 + t) \left(\frac {d}{d t} (t \cdot \sin t)\right) - t \cdot \sin t \left(\frac {d}{d t} (1 + t)\right)}{(1 + t) ^ {2}}$$ $$y ^ {\prime} (t) = \frac {(1 + t) \left(\frac {d}{d t} (t \cdot \sin t)\right) - t \cdot \sin t}{(1 + t) ^ {2}}$$

To find $\frac{d}{dt} (t*\sin t)$ we use the product rule $\frac{d}{dt} (uv) = v\frac{du}{dt} +u\frac{dv}{dt}$ , where $u = t$ and $v = \sin t$ :

$$y ^ {\prime} (t) = \frac {(1 + t) \left(t \cdot \frac {d}{d t} (\sin t) + \sin t \cdot \frac {d}{d t} (t)\right) - t \cdot \sin t}{(1 + t) ^ {2}}$$ $$y ^ {\prime} (t) = \frac {(1 + t) (t \cdot \cos t + \sin t) - t \cdot \sin t}{(1 + t) ^ {2}}$$

So, we can open the brackets:

$$y ^ {\prime} (t) = \frac {(1 + t) (t \cdot \cos t + \sin t) - t \cdot \sin t}{(1 + t) ^ {2}}$$ $$y ^ {\prime} (t) = \frac {t \cdot \cos t + \sin t + t ^ {2} \cdot \cos t + t \cdot \sin t - t \cdot \sin t}{(1 + t) ^ {2}}$$

Answer: the derivative of $y = \frac{\mathrm{t} \cdot \sin t}{1 + t}$ is

$$y ^ {\prime} (t) = \frac {\sin t + t \cdot (1 + t) \cdot \cos t}{(1 + t) ^ {2}}$$