Find fracdydx for: 1. Y = 2u3, u = 8x - 1 2. Y = sin u, u = x - xcos x | Basic Mathematics 2077

Basic Mathematics20775 marksNumerical: chain rule derivativesChain rule

Find $\frac{dy}{dx}$ for:

$Y = 2u^3$ , $u = 8x - 1$
$Y = \sin u$ , $u = x - x\cos x$

[5]

Find $\frac{dy}{dx}$ for:

$Y = 2u^3$, $u = 8x - 1$
$Y = \sin u$, $u = x - x\cos x$

[5]

Problem 1: $Y = 2u^3$ , $u = 8x - 1$ Problem 2: $Y = \sin u$ , $u = x - x\cos x$ Chain rule: $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ --- Step 1: Differentiate $Y$ with respect to $u$ $\frac{dy}{du} = 2 \cdot 3u^2 = 6u^2$ Step 2: Differentiate $u$ with respect to $x$ $$\frac{du}{dx}...

Problem 1: $Y = 2u^3$, $u = 8x - 1$ Problem 2: $Y = \sin u$, $u = x - x\cos x$ Chain rule: $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ --- Step 1: Differentiate $Y$ with respect to $u$ $$\frac{dy}{du} = 2 \cdot 3u^2 = 6u^2$$ Step 2: Differentiate $u$ with respect to $x$ $$\frac{du}{dx}...

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