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6. (a) Differentiate with respect to $x$, (i) $e^{3x}( ext{sin} x + 2 ext{cos} x)$, (ii) $x^3 ext{ln}(5x + 2)$ - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 5
Question 1
6. (a) Differentiate with respect to $x$, (i) $e^{3x}( ext{sin} x + 2 ext{cos} x)$, (ii) $x^3 ext{ln}(5x + 2)$. Given that $y = \frac{3x^2 + 6x - 7}{(x + 1)... show full transcript
Worked Solution & Example Answer:6. (a) Differentiate with respect to $x$, (i) $e^{3x}( ext{sin} x + 2 ext{cos} x)$, (ii) $x^3 ext{ln}(5x + 2)$ - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 5
Step 1
Differentiate with respect to $x$, (i) $e^{3x}( ext{sin} x + 2 ext{cos} x)$
Answer
To differentiate this expression, we will use the product rule, which states that if you have two functions multiplied together, ( u \cdot v ), then ( \frac{d}{dx}(uv) = u'v + uv' ). Here, let ( u = e^{3x} ) and ( v = ext{sin}(x) + 2 ext{cos}(x) ).
First, we differentiate ( u ): [ u = 3e^{3x} ] Next, we differentiate ( v ): [ \frac{d}{dx}(\text{sin}(x) + 2\text{cos}(x)) = \text{cos}(x) - 2\text{sin}(x) ] Applying the product rule: [ \frac{d}{dx}(e^{3x}(\text{sin} x + 2\text{cos} x)) = 3e^{3x}(\text{sin} x + 2\text{cos} x) + e^{3x}(\text{cos} x - 2\text{sin} x) ] This simplifies to: [ e^{3x}(3(\text{sin} x + 2\text{cos} x) + (\text{cos} x - 2\text{sin} x)) ]
Step 2
Differentiate with respect to $x$, (ii) $x^3 ext{ln}(5x + 2)$
Answer
For this part, we will again use the product rule. Let ( u = x^3 ) and ( v = \text{ln}(5x + 2) ).
First, differentiate ( u ): [ \frac{du}{dx} = 3x^2 ] Next, we differentiate ( v ) using the chain rule: [ \frac{dv}{dx} = \frac{1}{5x + 2} \cdot 5 = \frac{5}{5x + 2} ] Now applying the product rule: [ \frac{d}{dx}(x^3 \text{ln}(5x + 2)) = 3x^2 \text{ln}(5x + 2) + x^3 \cdot \frac{5}{5x + 2} ]
Step 3
show that $\frac{dy}{dx} = \frac{20}{(x + 1)^3}$
Answer
To show this, we need to differentiate the given function: [ y = \frac{3x^2 + 6x - 7}{(x + 1)^2} ] Using the quotient rule: [ \frac{dy}{dx} = \frac{(u'v - uv')}{v^2} ] where ( u = 3x^2 + 6x - 7 ) and ( v = (x + 1)^2 ).
Now, calculate ( u' ) and ( v' ): [ u' = 6x + 6, , v' = 2(x + 1) ] Substituting: [ \frac{dy}{dx} = \frac{(6x + 6)(x + 1)^2 - (3x^2 + 6x - 7)(2(x + 1))}{(x + 1)^4} ] Expanding and simplifying gives: [ = \frac{20}{(x + 1)^3} ]
Step 4
Hence find $\frac{d^2y}{dx^2}$ and the real values of $x$ for which $\frac{dy}{dx} = \frac{15}{4}$
Answer
From our earlier result: [ \frac{dy}{dx} = \frac{20}{(x + 1)^3} ] To find , we differentiate ( \frac{dy}{dx} ). [ \frac{d^2y}{dx^2} = -\frac{60}{(x + 1)^4} ] Now solving ( \frac{20}{(x + 1)^3} = \frac{15}{4} ): [ 20 \cdot 4 = 15(x + 1)^3 ] [ 80 = 15(x + 1)^3 ] Divide both sides by 15: [ \frac{16}{3} = (x + 1)^3 ] Taking the cube root: [ \sqrt[3]{\frac{16}{3}} = x + 1 ] Thus: [ x = \sqrt[3]{\frac{16}{3}} - 1 ]