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Given the following functions: $$f(x) = rown{x}$$ g(x) = 2x + 3 Let \( h(x) = f(g(x)) \) Find \( h'(x) \) h'(x) = (Total for Question 22 is 3 marks) - Edexcel - GCSE Maths - Question 22 - 2022 - Paper 2

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Question 22

Given-the-following-functions:--$$f(x)-=--rown{x}$$-g(x)-=-2x-+-3--Let-\(-h(x)-=-f(g(x))-\)--Find-\(-h'(x)-\)--h'(x)-=---(Total-for-Question-22-is-3-marks)-Edexcel-GCSE Maths-Question 22-2022-Paper 2.png

Given the following functions: $$f(x) = rown{x}$$ g(x) = 2x + 3 Let \( h(x) = f(g(x)) \) Find \( h'(x) \) h'(x) = (Total for Question 22 is 3 marks)

Worked Solution & Example Answer:Given the following functions: $$f(x) = rown{x}$$ g(x) = 2x + 3 Let \( h(x) = f(g(x)) \) Find \( h'(x) \) h'(x) = (Total for Question 22 is 3 marks) - Edexcel - GCSE Maths - Question 22 - 2022 - Paper 2

Step 1

Find \( g'(x) \)

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Answer

To find ( g'(x) ), we differentiate ( g(x) = 2x + 3 ):

g′(x)=2g'(x) = 2

Step 2

Find \( f'(x) \)

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Answer

To find ( f'(x) ), we differentiate ( f(x) = \sqrt{x} ) using the power rule:

f′(x)=12x−12=12xf'(x) = \frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}

Step 3

Apply the Chain Rule to find \( h'(x) \)

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101 rated

Answer

Using the Chain Rule, we have:

h′(x)=f′(g(x))⋅g′(x)h'(x) = f'(g(x)) \cdot g'(x)

Substituting in our expressions:

h′(x)=(12g(x))⋅g′(x)h'(x) = \left( \frac{1}{2\sqrt{g(x)}} \right) \cdot g'(x)

Now plug in ( g(x) = 2x + 3 ):

h′(x)=(122x+3)⋅2=12x+3h'(x) = \left( \frac{1}{2\sqrt{2x + 3}} \right) \cdot 2 = \frac{1}{\sqrt{2x + 3}}

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