Find y' for y = 3^((sin ax)/(cos bx)) + 1/3

Find $y'$ for $y = 3^{\frac{\sin ax}{\cos bx}} + \dfrac{1}{3}\cdot\dfrac{\sin^{3}ax}{\cos^{3}bx}$, where $a$, $b$ are constants.

Step 1: Recognise the common inner function

Set $T = \dfrac{\sin ax}{\cos bx}$. Then $\dfrac{\sin^{3}ax}{\cos^{3}bx} = T^{3}$, so the whole function is

\[ y = 3^{T} + \frac{T^{3}}{3}. \]

Both terms therefore differentiate through the same factor $T'$.

Step 2: Differentiate with respect to $T$

\[ \frac{dy}{dT} = 3^{T}\ln 3 + \frac{3T^{2}}{3} = 3^{T}\ln 3 + T^{2}. \]

Step 3: Differentiate the inner quotient

By the quotient rule, with $\bigl(\sin ax\bigr)' = a\cos ax$ and $\bigl(\cos bx\bigr)' = -b\sin bx$: \[T' = \frac{a\cos ax\cos bx - \sin ax\bigl(-b\sin bx\bigr)}{\cos^{2}bx} = \frac{a\cos ax\cos bx + b\sin ax\sin bx}{\cos^{2}bx}.\]

Step 4: Multiply

\[ y' = \left(3^{T}\ln 3 + T^{2}\right)T'. \]

Step 5: Verification

\[x = 0:\quad T = 0,\ T' = a \;\Longrightarrow\; y' = \bigl(1\cdot\ln 3 + 0\bigr)a = a\ln 3\]

\[a = b = 1,\ x = 0:\quad y' = \ln 3 \approx 1.09861; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 1.09861 \;\;\checkmark\]