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Exercise 10.5 · Q3

Q.If y=14u4y = \dfrac14 u^4, u=23x3+5u = \dfrac23 x^3+5, then dydx\dfrac{dy}{dx} is

(1) 127x2(2x3+15)3\dfrac{1}{27}x^2(2x^3+15)^3
(2) 227x(2x3+5)3\dfrac{2}{27}x(2x^3+5)^3
(3) 227x2(2x3+15)3\dfrac{2}{27}x^2(2x^3+15)^3
(4) −227x(2x3+5)3-\dfrac{2}{27}x(2x^3+5)^3
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✓ Free question

Step 1. y=14u4⇒dydu=u3y=\dfrac14u^4 \Rightarrow \dfrac{dy}{du}=u^3.

Step 2. u=23x3+5⇒dudx=2x2u=\dfrac23x^3+5 \Rightarrow \dfrac{du}{dx}=2x^2.

Step 3. By the chain rule:

dydx=dydu⋅dudx=u3⋅2x2\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=u^3\cdot 2x^2

Step 4. Write uu over a common denominator: u=2x3+153u=\dfrac{2x^3+15}{3}, so u3=(2x3+15)327u^3=\dfrac{(2x^3+15)^3}{27}.

Step 5. Substitute back:

dydx=(2x3+15)327⋅2x2=227x2(2x3+15)3\frac{dy}{dx}=\frac{(2x^3+15)^3}{27}\cdot 2x^2=\frac{2}{27}x^2(2x^3+15)^3

Step 6. Matching against the options, this is option (3) — note it uses (2x3+15)3(2x^3+15)^3 with x2x^2, not xx.

✓Final answer

The correct option is (3) 227x2(2x3+15)3\dfrac{2}{27}x^2(2x^3+15)^3

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