y = sin(3·x + 6)\nfind \\frac{dy}{dx}\n\\frac{dy}{dx} = \ntype sin(x) for sin(x), cos(x) for cos(x), and so…

y = sin(3·x + 6)\nfind \\frac{dy}{dx}\n\\frac{dy}{dx} = \ntype sin(x) for sin(x), cos(x) for cos(x), and so on.\nuse x^2 to square x, x^3 to cube x, and so on.\nuse ( sin(x) )^2 to square sin(x).\ndo not simplify your answer.\nquestion help: message instructor

y = sin(3·x + 6)\nfind \\frac{dy}{dx}\n\\frac{dy}{dx} = \ntype sin(x) for sin(x), cos(x) for cos(x), and so on.\nuse x^2 to square x, x^3 to cube x, and so on.\nuse ( sin(x) )^2 to square sin(x).\ndo not simplify your answer.\nquestion help: message instructor

Answer

Answer:

$3\cos(3x + 6)$

Explanation:

Step1: Let $u = 3x+6$

$y=\sin(u)$

Step2: Differentiate $y$ with respect to $u$

$\frac{dy}{du}=\cos(u)$

Step3: Differentiate $u$ with respect to $x$

$\frac{du}{dx}=3$

Step4: Use the chain rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

$\frac{dy}{dx}=\cos(u)\cdot3$

Step5: Substitute $u = 3x + 6$ back in

$\frac{dy}{dx}=3\cos(3x + 6)$