If y = csc(u), dy/dx equals?

If y = csc(u), dy/dx equals?

• A. -u' csc(u) cot(u)
• B. u' csc(u) cot(u)
• C. -u' sec(u) tan(u)
• D. -u' cos(u)/sin^2(u)

Answer: (A)

Recognize the chain rule at work: when y = csc(u) and u is a function of x, the derivative is dy/dx = (dy/du) · du/dx. The derivative of csc(u) with respect to u is -csc(u) cot(u). Multiply by du/dx (denoted u') to get dy/dx = -u' csc(u) cot(u). This matches the form given, since csc(u) cot(u) equals cos(u)/sin^2(u), so dy/dx can also be written as -u' cos(u)/sin^2(u). The negative sign and the csc cot combination are the key ideas here; common missteps would be using a positive sign or a different trig function, which don’t fit this derivative.

Recognize the chain rule at work: when y = csc(u) and u is a function of x, the derivative is dy/dx = (dy/du) · du/dx. The derivative of csc(u) with respect to u is -csc(u) cot(u). Multiply by du/dx (denoted u') to get dy/dx = -u' csc(u) cot(u). This matches the form given, since csc(u) cot(u) equals cos(u)/sin^2(u), so dy/dx can also be written as -u' cos(u)/sin^2(u). The negative sign and the csc cot combination are the key ideas here; common missteps would be using a positive sign or a different trig function, which don’t fit this derivative.