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5. (i) Differentiate with respect to x (a) y = x² ln 2x (b) y = (x + sin 2x)³ Given that x = cot y, (ii) show that \( \frac{dy}{dx} = \frac{-1}{1+x^2} \) - Edexcel - A-Level Maths Pure - Question 25 - 2013 - Paper 1
Question 25
5. (i) Differentiate with respect to x (a) y = x² ln 2x (b) y = (x + sin 2x)³ Given that x = cot y, (ii) show that \( \frac{dy}{dx} = \frac{-1}{1+x^2} \)
Worked Solution & Example Answer:5. (i) Differentiate with respect to x (a) y = x² ln 2x (b) y = (x + sin 2x)³ Given that x = cot y, (ii) show that \( \frac{dy}{dx} = \frac{-1}{1+x^2} \) - Edexcel - A-Level Maths Pure - Question 25 - 2013 - Paper 1
Step 1
Differentiate with respect to x (a)
Answer
To differentiate the expression ( y = x^2 \ln 2x ), we will apply the product rule, which states that ( (uv)' = u'v + uv' ).
Let ( u = x^2 ) and ( v = \ln 2x ).
Calculating the derivatives:
- ( u' = 2x )
- To find ( v' ), we use the chain rule: ( \ln 2x = \ln 2 + \ln x ), so ( v' = \frac{1}{x} ).
Now applying the product rule: [ \frac{dy}{dx} = 2x \ln 2x + x^2 \cdot \frac{1}{x} = 2x \ln 2x + x. ]
Step 2
Differentiate with respect to x (b)
Answer
To differentiate the expression ( y = (x + \sin 2x)^3 ), we apply the chain rule.
Letting ( u = x + \sin 2x ), the derivative is given by: [ \frac{dy}{dx} = 3u^2 \cdot \frac{du}{dx}. ]
Now we calculate ( \frac{du}{dx} ):
- The derivative of ( u = x + \sin 2x ) is ( \frac{du}{dx} = 1 + 2\cos 2x ).
Putting it all together: [ \frac{dy}{dx} = 3(x + \sin 2x)^2 (1 + 2\cos 2x). ]
Step 3
show that \( \frac{dy}{dx} = \frac{-1}{1+x^2} \) (ii)
Answer
Given that ( x = \cot y ), we can use implicit differentiation. From ( x = \cot y ), we know: [ \frac{dx}{dy} = -\csc^2 y. ]
Using the relationship between derivatives: [ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{-1}{\csc^2 y} = \frac{-1}{1 + \cot^2 y} = \frac{-1}{1+x^2}. ]
Thus, this shows the required expression.