Answer:
[tex]\frac{dy}{dx} = 2xcos(x^2+7)[/tex]
Step-by-step explanation:
Given: [tex]y=sin(x^2+7)[/tex]
Find: [tex]\frac{dy}{dx}[/tex]
For this problem, it looks like we are going to use the chain rule. We have to functions, let's label them:
[tex]f(x)=sin(x)\\g(x)=x^2+7[/tex]
Let's remember that the chain rule is: [tex]f'(g(x))*g'(x)[/tex] or [tex](\frac{dy}{du} )(\frac{du}{dx} )[/tex]
Since these functions are basic, I will be using the [tex]f'(g(x))*g'(x)[/tex] rule, but both rules always work. We need to find [tex]f'(x)[/tex] and [tex]g'(x)[/tex]
[tex]f'(x)=cos(x)\\g'(x)=2x[/tex]
Remember that the derivative of [tex]sin(x)[/tex] is [tex]cos(x)[/tex]. Make sure you know all your trig derivatives. Now that we know all our variables, we can plug them in. We will take [tex]g(x)[/tex] and plug it into [tex]f'(x)[/tex] and multiply it all by [tex]g'(x)[/tex].
[tex][cos(x^2+7)][2x]=2xcos(x^2+7)[/tex]
Since nothing else can be done to reduce this expression, your answer is:
[tex]\frac{dy}{dx} = 2xcos(x^2+7)[/tex]
