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The functions g and h are such that g(x) = √(2x - 5) h(x) = 1/x (a) Find g(16) (b) Find hg'(x) Give your answer in terms of x in its simplest form - Edexcel - GCSE Maths - Question 19 - 2022 - Paper 2
Question 19
The functions g and h are such that g(x) = √(2x - 5) h(x) = 1/x (a) Find g(16) (b) Find hg'(x) Give your answer in terms of x in its simplest form. hg'(x) =
Worked Solution & Example Answer:The functions g and h are such that g(x) = √(2x - 5) h(x) = 1/x (a) Find g(16) (b) Find hg'(x) Give your answer in terms of x in its simplest form - Edexcel - GCSE Maths - Question 19 - 2022 - Paper 2
Step 1
Step 2
Find hg'(x)
Answer
To find hg'(x), we first need to compute g'(x). The function g(x) is:
g(x) = ext{√(2x - 5)}$$ Using the chain rule, we get:g'(x) = rac{1}{2 ext{√(2x - 5)}} imes 2 = rac{1}{ ext{√(2x - 5)}}$$
Now, we find h(g(x)):
h(g(x)) = rac{1}{g(x)} = rac{1}{ ext{√(2x - 5)}}$$ Next, we take the derivative of hg(x): Using the quotient rule:hg'(x) = rac{-g'(x)}{(g(x))^2}$$
Substituting g'(x) and g(x):
hg'(x) = - rac{ rac{1}{ ext{√(2x - 5)}}}{ ext{(√(2x - 5))}^2} = - rac{1}{(2x - 5)√(2x - 5)}$$ Thus, the final answer is:hg'(x) = - rac{1}{(2x - 5)√(2x - 5)}$$