Question

Find \(d y / d x\) if \(y=\cos (6 x+2)\) by writing \(y\) as a composite with \(\begin{array}{ll}{\text { (a) } y=\cos u} & {\text { and } u=6 x+2} \\\ {\text { (b) } y=\cos 2 u} & {\text { and } \quad u=3 x+1}\end{array}\)

Short answer

The derivative of the function \(y = \cos(6x + 2)\) is \(-6\sin(6x + 2)\) whether written as composite function (a) or (b)

Step-by-step solution

Find the derivative using first composite function

We are given that \(y = \cos u\) and \(u = 6x + 2\). The chain rule states that \(dy/dx = dy/du * du/dx\). So, differentiating \(y\) with respect to \(u\) gives \(-\sin u\), and differentiating \(u\) with respect to \(x\) gives 6. Therefore, \(dy/dx = -\sin u * 6 = -6\sin u\).

Substitute \(u\) back into the function

Replace \(u\) with \(6x + 2\) to write the derivative in terms of \(x\). So, \(dy/dx = -6\sin(6x + 2)\). This is the derivative of the given function when \(y\) is written with composite function (a).

Find the derivative using second composite function

Now, we are given \(y = \cos 2u\) and \(u= 3x + 1\). Applying the chain rule again, we first differentiate \(y\) with respect to \(u\) to get \(-2\sin 2u\) and differentiate \(u\) with respect to \(x\) to get 3. So, \(dy/dx = -2\sin 2u * 3 = -6\sin 2u\).

Substitute \(u\) back into the function

Again, replace \(u\) with \(3x + 1\) to write the derivative in terms of \(x\). So, \(dy/dx = -6\sin(2(3x + 1)) = -6\sin(6x + 2)\). This is the derivative of the given function when \(y\) is written with composite function (b).