Differentiate $W=e^{9 y}+\frac{1}{e^{9 y}}$
Answer: $\frac{d W}{d y}=$

Step 1 ：We are given the function \(W=e^{9 y}+\frac{1}{e^{9 y}}\). We are asked to find its derivative with respect to \(y\).

Step 2 ：To differentiate this function, we need to apply the chain rule and the rule for the derivative of the exponential function.

Step 3 ：The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

Step 4 ：The derivative of \(e^x\) is \(e^x\). The derivative of \(e^{-x}\) is \(-e^{-x}\).

Step 5 ：Therefore, the derivative of \(e^{9y}\) is \(9e^{9y}\) and the derivative of \(\frac{1}{e^{9y}}\) is \(-9e^{-9y}\).

Step 6 ：So, the derivative of the function \(W=e^{9 y}+\frac{1}{e^{9 y}}\) with respect to \(y\) is \(9e^{9y} - 9e^{-9y}\).

Step 7 ：Final Answer: \(\frac{d W}{d y}=\boxed{9e^{9y} - 9e^{-9y}}\)