a) Differentiate $ \frac{1}{3} x^{2} - x + 3 $ from first principles with respect to $ x $. b) $ f(x) = \ln(3x^{2} + 2) $ and $ g(x) = x + 5 $, where $ x \in \mathbb{R} $. Find the value of the derivative of $ f(g(x)) $ at $ x = -\frac{1}{4} $. Give your answer correct to 3 decimal places.

Leaving Cert Mathematics Differentiation: a) Differentiate \( \frac{1}{3}

a) Differentiate $ \frac{1}{3} x^{2} - x + 3 $ from first principles with respect to $ x $ - Leaving Cert Mathematics - Question 3 - 2017

Question 3

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a) Differentiate $ \frac{1}{3} x^{2} - x + 3 $ from first principles with respect to $ x $. b) $ f(x) = \ln(3x^{2} + 2) $ and $ g(x) = x + 5 $, where \( x... show full transcript

Worked Solution & Example Answer:a) Differentiate $ \frac{1}{3} x^{2} - x + 3 $ from first principles with respect to $ x $ - Leaving Cert Mathematics - Question 3 - 2017

Step 1

Differentiate $ \frac{1}{3} x^{2} - x + 3 $ from first principles

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Answer

To differentiate ( f(x) = \frac{1}{3} x^{2} - x + 3 ) from first principles, we apply the limit definition of the derivative:

df(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Calculating ( f(x+h) ):
( f(x+h) = \frac{1}{3} (x+h)^{2} - (x+h) + 3 )
( = \frac{1}{3} (x^{2} + 2xh + h^{2}) - x - h + 3 )
( = \frac{1}{3} x^{2} + \frac{2}{3}xh + \frac{1}{3}h^{2} - x - h + 3 )
( = \frac{1}{3} x^{2} - x + 3 + \frac{2}{3}xh - h + \frac{1}{3}h^{2} )

Now, substituting back into the limit definition:

\lim_{h \to 0} \frac{\left( \frac{1}{3} x^{2} - x + 3 + \frac{2}{3}xh - h + \frac{1}{3}h^{2} \right) - \left( \frac{1}{3} x^{2} - x + 3 \right)}{h} \ = \lim_{h \to 0} \frac{\frac{2}{3}xh - h + \frac{1}{3}h^{2}}{h} \

This simplifies to:

= \lim_{h \to 0} \left( \frac{2}{3}x - 1 + \frac{1}{3}h \right) = \frac{2}{3}x - 1

Therefore, the derivative is ( f'(x) = \frac{2}{3}x - 1 ).

Step 2

Find the value of the derivative of $ f(g(x)) $ at $ x = -\frac{1}{4} $

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Answer

We need to use the chain rule to find the derivative of ( f(g(x)) ):

Using the chain rule:

d(f(g(x))) = f'(g(x)) \cdot g'(x)

Where:
( f'(x) = \frac{1}{3x^{2} + 2} \cdot (6x) ) (found from differentiating ( f(x) )).
( g'(x) = 1 ) (since ( g(x) = x + 5 )).

First, calculate ( g(-\frac{1}{4}) ):
( g(-\frac{1}{4}) = -\frac{1}{4} + 5 = \frac{19}{4} )

Now calculate ( f'(g(-\frac{1}{4})) ):
( f' \left( \frac{19}{4} \right) = \ln\left( 3\left(\frac{19}{4}\right)^{2} + 2 \right) )

Substituting this into the chain rule gives:

d(f(g(-\frac{1}{4}))) = f' \left( \frac{19}{4} \right) \cdot 1 \approx 0.372

Thus, the value of the derivative at ( x = -\frac{1}{4} ) is approximately ( 0.372 ) correct to three decimal places.

a) Differentiate \( \frac{1}{3} x^{2} - x + 3 \) from first principles with respect to \( x \) - Leaving Cert Mathematics - Question 3 - 2017

Worked Solution & Example Answer:a) Differentiate \( \frac{1}{3} x^{2} - x + 3 \) from first principles with respect to \( x \) - Leaving Cert Mathematics - Question 3 - 2017

Differentiate \( \frac{1}{3} x^{2} - x + 3 \) from first principles

Find the value of the derivative of \( f(g(x)) \) at \( x = -\frac{1}{4} \)

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