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junior college 1 | H1 Maths
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Date Posted: 4 years ago
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I am assuming that this is a differentiation question. We simplify the fraction first before performing the differentiation. We can also use the quotient rule in this case, but simplification of the expression is preferred.
Remember that chain rule continues to apply.
Also do note that we CANNOT differentiate the numerator and the denominator separately before dividing the result together.
Remember that chain rule continues to apply.
Also do note that we CANNOT differentiate the numerator and the denominator separately before dividing the result together.
Date Posted:
4 years ago
Thanks for the clarification!
Quotient rule way :
d/dx (3 - e^2x)² / e^x
= [ e^x (2)(3 - e^(2x))(-2e^(2x)) - (3 - e^(2x))² e^x ] / (e^x)²
= [ 2(3 - e^(2x))(-2e^(2x)) - (3 - e^(2x))² ] / e^x
= [ (3 - e^(2x)) (-4e^(2x) - (3 - e^(2x)) ] / e^x
= (3 - e^(2x)) (-3e^(2x) - 3) / e^x
= (-9e^(2x) - 9 + 3e^(4x) + 3e^(2x) ) / e^x
= (3e^(4x) - 6 e^(2x) - 9) / e^x
= 3e^(3x) - 6e^x - 9/e^x
d/dx (3 - e^2x)² / e^x
= [ e^x (2)(3 - e^(2x))(-2e^(2x)) - (3 - e^(2x))² e^x ] / (e^x)²
= [ 2(3 - e^(2x))(-2e^(2x)) - (3 - e^(2x))² ] / e^x
= [ (3 - e^(2x)) (-4e^(2x) - (3 - e^(2x)) ] / e^x
= (3 - e^(2x)) (-3e^(2x) - 3) / e^x
= (-9e^(2x) - 9 + 3e^(4x) + 3e^(2x) ) / e^x
= (3e^(4x) - 6 e^(2x) - 9) / e^x
= 3e^(3x) - 6e^x - 9/e^x
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