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

Step 1 ：We are given the function $W=e^{9y}+\frac{1}{e^{9y}}$. 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 $W=e^{9y}+\frac{1}{e^{9y}}$ with respect to $y$ is $9e^{9y}-9e^{-9y}$.

Step 7 ：Final Answer: $\frac{dW}{dy}=\boxed{{\displaystyle 9e^{9y}-9e^{-9y}}}$