find the derivative of $z = qcdot e^{cos q}$\n$\frac{dz}{dq}=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $z = u\cdot v$, where $u = q$ and $v=e^{\cos q}$, then $\frac{dz}{dq}=u'\cdot v + u\cdot v'$. First, find $u'$. Since $u = q$, then $u'=\frac{d}{dq}(q)=1$.

Step2: Find $v'$

Let $t=\cos q$. Then $v = e^{t}$. By the chain - rule, $\frac{dv}{dq}=\frac{dv}{dt}\cdot\frac{dt}{dq}$. We know that $\frac{dv}{dt}=\frac{d}{dt}(e^{t}) = e^{t}$ and $\frac{dt}{dq}=\frac{d}{dq}(\cos q)=-\sin q$. So $v'=\frac{dv}{dq}=e^{\cos q}\cdot(-\sin q)=-e^{\cos q}\sin q$.

Step3: Calculate $\frac{dz}{dq}$

Substitute $u = q$, $u' = 1$, $v = e^{\cos q}$, and $v'=-e^{\cos q}\sin q$ into the product - rule formula: $\frac{dz}{dq}=1\cdot e^{\cos q}+q\cdot(-e^{\cos q}\sin q)=e^{\cos q}(1 - q\sin q)$.

Answer:

$e^{\cos q}(1 - q\sin q)$