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4. (a) Differentiate with respect to x (i) $x^2 e^{x^2}$ - Edexcel - A-Level Maths Pure - Question 5 - 2006 - Paper 5
Question 5
4. (a) Differentiate with respect to x (i) $x^2 e^{x^2}$. (ii) $\frac{\cos(2x)}{3x}$. (b) Given that $x = 4 \sin(2y + 6)$, find $\frac{dy}{dx}$ in terms of x.
Worked Solution & Example Answer:4. (a) Differentiate with respect to x (i) $x^2 e^{x^2}$ - Edexcel - A-Level Maths Pure - Question 5 - 2006 - Paper 5
Step 1
(i) $x^2 e^{x^2}$
Answer
To differentiate the function (y = x^2 e^{x^2}), we will use the product rule. The product rule states that if you have a function (uv), then the derivative is given by (u'v + uv').
Let:
- (u = x^2) and (v = e^{x^2})
- Differentiate (u:\ u' = 2x)
- Differentiate (v:\ v' = e^{x^2} \cdot 2x) (using the chain rule)
Now apply the product rule: [ \frac{dy}{dx} = u'v + uv' = (2x)e^{x^2} + (x^2)(2xe^{x^2}) = 2xe^{x^2} + 2x^3e^{x^2} = 2xe^{x^2}(1 + x^2) ]
Step 2
(ii) $\frac{\cos(2x)}{3x}$
Answer
To differentiate (y = \frac{\cos(2x)}{3x}), we will use the quotient rule, which states that if we have a function (\frac{u}{v}), then [ \frac{dy}{dx} = \frac{u'v - uv'}{v^2}. ] Let:
- (u = \cos(2x)) and (v = 3x)
- Differentiate (u:\ u' = -2 \sin(2x)) (using the chain rule)
- Differentiate (v:\ v' = 3)
Now apply the quotient rule: [ \frac{dy}{dx} = \frac{(-2 \sin(2x))(3x) - (\cos(2x))(3)}{(3x)^2} = \frac{-6x \sin(2x) - 3 \cos(2x)}{9x^2} = \frac{-2(2x \sin(2x) + \cos(2x))}{9x^2}. ]
Step 3
(b) Given that $x = 4 \sin(2y + 6)$, find $\frac{dy}{dx}$ in terms of x.
Answer
To find (\frac{dy}{dx}), we first differentiate both sides of the equation with respect to (x).
Given: [x = 4 \sin(2y + 6)] We will use implicit differentiation: [\frac{dx}{dx} = 4 \cdot \cos(2y + 6) \cdot \frac{d(2y + 6)}{dx}]
Applying the chain rule: [1 = 4 \cos(2y + 6) \cdot 2 \frac{dy}{dx}]
Now solving for (\frac{dy}{dx}): [ \frac{dy}{dx} = \frac{1}{8 \cos(2y + 6)}.]
Now we have (\frac{dy}{dx}) in terms of y. To express it in terms of x, we need to find (y) in terms of (x) from our original equation. From (x = 4 \sin(2y + 6)), we can write: [2y + 6 = \arcsin\left(\frac{x}{4}\right)].
Thus: [2y = \arcsin\left(\frac{x}{4}\right) - 6 \text{, hence } y = \frac{1}{2}\arcsin\left(\frac{x}{4}\right) - 3.]
Substituting y back: [ \frac{dy}{dx} = \frac{1}{8 \cos(\arcsin(\frac{x}{4}))}.]
Since (\cos(\arcsin(z)) = \sqrt{1 - z^2}), we find: [\cos(\arcsin(\frac{x}{4})) = \sqrt{1 - \left(\frac{x}{4}\right)^2} = \sqrt{1 - \frac{x^2}{16}}.]
Finally, the derivative becomes: [ \frac{dy}{dx} = \frac{1}{8 \sqrt{1 - \frac{x^2}{16}}}.]