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secondary 3 | A Maths
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how to do C！！helpp！thankyou！
Date Posted: 4 years ago
Views: 351
(1 + 2x)/√x
= 1/√x + 2x/√x
= 1/x¹⁄² + 2x/x¹⁄²
= x-¹⁄² + 2x¹-¹⁄²
= x-¹⁄² + 2x¹⁄²
So,
d/dx (1 + 2x)/√x
= d/dx (x-¹⁄² + 2x¹⁄²)
= -½x-¹⁄² -¹ + (½)(2x¹⁄²-¹)
= -½x-³⁄² + x-¹⁄²
= 1/x¹⁄² - 1/(2x³⁄²)
= 1/√x - 1/(2√x³)
Or if you want to leave it as 1 term,
Then
= -½x-³⁄² + x-³⁄²+¹
= -½x-³⁄² + x¹(x-³⁄²)
= (x - ½)(x-³⁄²)
= (x - ½)/x³⁄²
= (x - ½)/√x³
= (2x - 1)/(2√x³)
= 1/√x + 2x/√x
= 1/x¹⁄² + 2x/x¹⁄²
= x-¹⁄² + 2x¹-¹⁄²
= x-¹⁄² + 2x¹⁄²
So,
d/dx (1 + 2x)/√x
= d/dx (x-¹⁄² + 2x¹⁄²)
= -½x-¹⁄² -¹ + (½)(2x¹⁄²-¹)
= -½x-³⁄² + x-¹⁄²
= 1/x¹⁄² - 1/(2x³⁄²)
= 1/√x - 1/(2√x³)
Or if you want to leave it as 1 term,
Then
= -½x-³⁄² + x-³⁄²+¹
= -½x-³⁄² + x¹(x-³⁄²)
= (x - ½)(x-³⁄²)
= (x - ½)/x³⁄²
= (x - ½)/√x³
= (2x - 1)/(2√x³)
Alternatively, using product rule,
d/dx (1 + 2x)/√x
= d/dx (1 + 2x) (1/x¹⁄²)
= d/dx (1 + 2x)(x-¹⁄²)
= 2(x-¹⁄²) + (1 + 2x)(-½x-¹⁄²-¹)
= 2/x¹⁄² - ½x-³⁄² - x¹(x-³⁄²)
= 2/x¹⁄² - 1/(2x³⁄²) - x-¹⁄²
= 2/x¹⁄² - 1/(2x³⁄²) - 1/x¹⁄²
= 1/x¹⁄² - 1/(2x³⁄²)
= 1/√x - 1/(2√x³)
d/dx (1 + 2x)/√x
= d/dx (1 + 2x) (1/x¹⁄²)
= d/dx (1 + 2x)(x-¹⁄²)
= 2(x-¹⁄²) + (1 + 2x)(-½x-¹⁄²-¹)
= 2/x¹⁄² - ½x-³⁄² - x¹(x-³⁄²)
= 2/x¹⁄² - 1/(2x³⁄²) - x-¹⁄²
= 2/x¹⁄² - 1/(2x³⁄²) - 1/x¹⁄²
= 1/x¹⁄² - 1/(2x³⁄²)
= 1/√x - 1/(2√x³)
See 1 Answer
done
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clear
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Or you can directly use the quotient rule.
Date Posted:
4 years ago
The 2x√x would be simplified to 2√x³
True. Thanks
Most welcome Arnold.
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