So: ( 2x y^3 + 3x^2 y^2 \frac{dy}{dx} + \cos(y) \frac{dy}{dx} = 5 )
Then checked the solution in the back: — ( y = [\sin(4x)]^3 ) Let ( u = \sin(4x) ), then ( y = u^3 ), ( \frac{dy}{du} = 3u^2 ) ( \frac{du}{dx} = \cos(4x) \cdot 4 ) (chain rule again inside) ( \frac{dy}{dx} = 3[\sin(4x)]^2 \cdot 4\cos(4x) = 12\sin^2(4x)\cos(4x) ) ✓ She had gotten it right — but the solution reminded her to explicitly show the inner chain rule on (4x), a step she often rushed. A Week Later — The Improvement Mia did two chapters per night. On Wednesday, she tackled implicit differentiation : Problem 47 — Find ( \frac{dy}{dx} ) for ( x^2 y^3 + \sin(y) = 5x ) She wrote: So: ( 2x y^3 + 3x^2 y^2 \frac{dy}{dx}
[ \frac{d}{dx}[x^2 y^3] + \frac{d}{dx}[\sin(y)] = \frac{d}{dx}[5x] ] Definite Integral by u-Substitution Problem : Evaluate (
Using product rule on first term: ( 2x \cdot y^3 + x^2 \cdot 3y^2 \frac{dy}{dx} ) then ( y = u^3 )
: ( h'(x) = (e^{2x})' \cos(3x) + e^{2x} (\cos(3x))' ) ( = 2e^{2x} \cos(3x) + e^{2x} \cdot (-\sin(3x) \cdot 3) ) ( = e^{2x}[2\cos(3x) - 3\sin(3x)] ) 3. Definite Integral by u-Substitution Problem : Evaluate ( \int_{0}^{\pi/2} \sin x \cos^3 x , dx )
Right side: ( 5 )