Question

find $\frac{dy}{dx}$ for $y = - 19x + 3cos x$.
$\frac{d}{dx}(-19x + 3cos x)=square$

Explanation:

Step1: Apply sum - difference rule of derivatives

The derivative of a sum/difference of functions $u(x)\pm v(x)$ is $u'(x)\pm v'(x)$. Here $u(x)=- 19x$ and $v(x)=3\cos x$. So $\frac{d}{dx}(-19x + 3\cos x)=\frac{d}{dx}(-19x)+\frac{d}{dx}(3\cos x)$.

Step2: Differentiate $-19x$

Using the power - rule $\frac{d}{dx}(ax)=a$ for a constant $a$. For $y = - 19x$, $\frac{d}{dx}(-19x)=-19$.

Step3: Differentiate $3\cos x$

Using the constant - multiple rule $\frac{d}{dx}(cf(x))=c\frac{d}{dx}(f(x))$ and the derivative of $\cos x$ which is $-\sin x$. So $\frac{d}{dx}(3\cos x)=3\frac{d}{dx}(\cos x)=3\times(-\sin x)=-3\sin x$.

Step4: Combine the results

$\frac{d}{dx}(-19x)+\frac{d}{dx}(3\cos x)=-19-3\sin x$.

Answer:

$-19 - 3\sin x$