If y = (x^2 + 3x)cosx , find the value of dy/dx when x=0
Can someone show me how to do this q please.
Thanks
Quotient Rule? You'd be much better off with the Product Rule since we're not doing any division here.
The product rule is $\displaystyle \dfrac{dy}{dx} \left[u(x)v(x)\right] = v\dfrac{du}{dx} + u\dfrac{dv}{dx}$
In your case $\displaystyle u = x^2 +3x$ and $\displaystyle v = \cos(x)$
Once you've differentiated find $\displaystyle y'(0)$. For reference I get an answer of 3