Example Question - chain rule differentiation
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Derivative of a Composition of Functions using the Chain Rule
The given function is \( y = 3(x^2 - x)^3 \). To find \( \frac{dy}{dx} \), we need to differentiate this function with respect to \( x \). This is a composition of functions, and therefore we will need to apply the chain rule to differentiate it. The chain rule states that if we have a function \( h(x) = f(g(x)) \), then its derivative is \( h'(x) = f'(g(x)) \cdot g'(x) \). In your case, let \( u = x^2 - x \). Then the function \( y \) can be rewritten as \( y = 3u^3 \). Now, we will differentiate both sides with respect to \( x \): 1. Differentiate \( y \) with respect to \( u \): \[ \frac{dy}{du} = 3 \cdot 3u^2 = 9u^2 \] 2. Differentiate \( u = x^2 - x \) with respect to \( x \): \[ \frac{du}{dx} = 2x - 1 \] Now, apply the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 9u^2(2x - 1) \] Substitute \( u \) back in terms of \( x \): \[ \frac{dy}{dx} = 9(x^2 - x)^2(2x - 1) \] Remember to expand and simplify the expression if needed. But as per the instruction given in the question, you wanted the expression for \( \frac{dy}{dx} \), which is what we've found.
Differentiation of Implicit Function
The given equation is y^2 = 9x^2 - 4x. To find dy/dx, differentiate both sides of the equation with respect to x, applying the chain rule to the term involving y because y is a function of x. Differentiating y^2 with respect to x yields: 2y * dy/dx (using the chain rule because y is a function of x) Differentiating 9x^2 - 4x with respect to x yields: 18x - 4 Now, equate these two results: 2y * dy/dx = 18x - 4 To solve for dy/dx, we divide both sides by 2y: dy/dx = (18x - 4) / (2y) To simplify the expression, we divide each term in the numerator by 2: dy/dx = (9x - 2) / y Now we have dy/dx. To find d^2y/dx^2, we need to differentiate dy/dx with respect to x using the quotient rule. Let's call u = 9x - 2 and v = y. The quotient rule states that if you have a function g(x) = u/v, its derivative g'(x) is given by: g'(x) = (u'v - uv') / v^2 Differentiating u = 9x - 2 with respect to x gives us u' = 9. Differentiating v = y with respect to x gives us v' = dy/dx. So applying the quotient rule gives us: d^2y/dx^2 = (9 * y - (9x - 2) * dy/dx) / y^2 Since we've found dy/dx previously as (9x - 2) / y, we substitute that in for dy/dx in the above expression: d^2y/dx^2 = (9 * y - (9x - 2) * ((9x - 2) / y)) / y^2 Let's simplify this expression: d^2y/dx^2 = (9 * y^2 - (9x - 2) * (9x - 2)) / y^3 At this point, it would be best to expand the numerator and then simplify the expression for the second derivative. d^2y/dx^2 = (9y^2 - (81x^2 - 18x*2 + 2^2)) / y^3 d^2y/dx^2 = (9y^2 - (81x^2 - 36x + 4)) / y^3 d^2y/dx^2 = (9y^2 - 81x^2 + 36x - 4) / y^3 Now we have both the first and second derivatives: dy/dx = (9x - 2) / y d^2y/dx^2 = (9y^2 - 81x^2 + 36x - 4) / y^3
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