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Differentiate with respect to x (a) ln(x^2 + 3x + 5) (b) \frac{\cos x}{x^2} - Edexcel - A-Level Maths Pure - Question 3 - 2011 - Paper 3
Question 3
Differentiate with respect to x (a) ln(x^2 + 3x + 5) (b) \frac{\cos x}{x^2}
Worked Solution & Example Answer:Differentiate with respect to x (a) ln(x^2 + 3x + 5) (b) \frac{\cos x}{x^2} - Edexcel - A-Level Maths Pure - Question 3 - 2011 - Paper 3
Step 1
(a) ln(x^2 + 3x + 5)
Answer
To differentiate the function ( y = \ln(u) ) where ( u = x^2 + 3x + 5 ), we apply the chain rule:
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Differentiate ( y ) with respect to ( u ): [ \frac{dy}{du} = \frac{1}{u} = \frac{1}{x^2 + 3x + 5} ]
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Differentiate ( u ) with respect to ( x ): [ \frac{du}{dx} = 2x + 3 ]
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Applying the chain rule: [ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{x^2 + 3x + 5} \cdot (2x + 3) ] Hence, the derivative is: [ \frac{2x + 3}{x^2 + 3x + 5} ]
Step 2
(b) \frac{\cos x}{x^2}
Answer
To differentiate the function ( y = \frac{u}{v} ) where ( u = \cos x ) and ( v = x^2 ), we use the quotient rule:
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The quotient rule states that: [ \frac{dy}{dx} = \frac{v u' - u v'}{v^2} ]
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We calculate the derivatives:
- ( u' = -\sin x )
- ( v' = 2x )
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Substitute into the quotient rule: [ \frac{dy}{dx} = \frac{x^2(-\sin x) - \cos x(2x)}{(x^2)^2} ] [ = \frac{-x^2 \sin x - 2x \cos x}{x^4} ]
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Simplifying gives: [ = \frac{-x \sin x - 2 \cos x}{x^3} ] and therefore the derivative is: [ \frac{-x \sin x - 2 \cos x}{x^3} ]