MATHEMATICS
JAMB 2005 - Question 37
Mathematics 2005 JAMB Past Questions - Question 37: Find the derivative of y = sin (2x³ + 3x -4)
A:
B:
C:
D:
Correct Answer
C
Explanation
To find the derivative of y = sin(2x³ + 3x - 4), we can use the chain rule. Let's denote the function inside the sine as u:u = 2x³ + 3x - 4Now, we can find the derivative of u with respect to x:du/dx = 6x² + 3Next, we can find the derivative of y with respect to u:dy/du = cos(u)Finally, we can apply the chain rule by multiplying the derivatives:dy/dx = dy/du * du/dx = cos(u) * (6x² + 3)Substituting back the value of u:dy/dx = cos(2x³ + 3x - 4) * (6x² + 3)Therefore, the derivative of y = sin(2x³ + 3x - 4) is dy/dx = cos(2x³ + 3x - 4) * (6x² + 3).
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