1. Problem: Find the differential coefficient (derivative) of $\tan(\sin(ax+b))$. 2. Formula and rules: Use the chain rule for derivatives. If $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$. 3. Step-by-step solution: - Let $y = \tan(\sin(ax+b))$. - Outer function: $f(u) = \tan u$, derivative $f'(u) = \sec^2 u$. - Inner function: $u = \sin(ax+b)$, derivative $u' = \cos(ax+b) \cdot a$ (chain rule again). 4. Applying chain rule: $$\frac{dy}{dx} = \sec^2(\sin(ax+b)) \cdot \cos(ax+b) \cdot a$$ 5. Final answer: $$\boxed{\frac{dy}{dx} = a \sec^2(\sin(ax+b)) \cos(ax+b)}$$