Question

Calculus Help Please!!!

Answer 1

Hello from MrBillDoesMath!

Answer:

y' = -1/(1+sinx)

the fourth choice.

Discussion:

Using the derivative quotient rule

y' = ( (1 + sinx) (-sinx) - cosx ( cosx) ) \ ( 1 + sinx)^2

as (cosx)' = -sinx and (sinx)' = cos

Expanding we get

y' = ( -sinx - (sinx)^2 - (cosx)^2 ) \ (1 + sinx)^2

But (sinx)^2 + (cos)^2 = 1 so this equals

y' = ( -sinx -1 )\ ( 1 + sinx)^2 =>

y' = - ( 1 + sinx) / (1 + sinx)^2

Notice the numerator is the square of the denominator so

y' = -1/(1+sinx)

which is the fourth choice

Thank you,

MrB

Answer 2

`\displaystyle y' = (-1)/(1 + \sin x)`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Addition/Subtraction]:
`\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]:
`\displaystyle (d)/(dx) [(f(x))/(g(x)) ]=(g(x)f'(x)-g'(x)f(x))/(g^2(x))`

Explanation:

Step 1: Define

Identify

`\displaystyle y = (\cos x)/(1 + \sin x)`

Step 2: Differentiate

1. [Function] Derivative Rule [Quotient Rule]:
   `\displaystyle y' = ((\cos x)'(1 + \sin x) - \cos x(1 + \sin x)')/((1 + \sin x)^2)`
2. Trigonometric Differentiation:
   `\displaystyle y' = (-\sin x(1 + \sin x) - \cos x(1 + \sin x)')/((1 + \sin x)^2)`
3. Rewrite [Derivative Rule - Addition/Subtraction]:
   `\displaystyle y' = (-\sin x(1 + \sin x) - \cos x[(1)' + (\sin x)'])/((1 + \sin x)^2)`
4. Trigonometric Differentiation:
   `\displaystyle y' = (-\sin x(1 + \sin x) - \cos x[(1)' + \cos x])/((1 + \sin x)^2)`
5. Basic Power Rule:
   `\displaystyle y' = (-\sin x(1 + \sin x) - \cos^2 x)/((1 + \sin x)^2)`
6. Factor:
   `\displaystyle y' = (- \big[ \sin x(1 + \sin x) + \cos^2 x \big])/((1 + \sin x)^2)`
7. Expand:
   `\displaystyle y' = (- \big[ \sin x+ \sin^2 x + \cos^2 x \big])/((1 + \sin x)^2)`
8. Simplify:
   `\displaystyle y' = (-(\sin x + 1))/((1 + \sin x)^2)`
9. Simplify:
   `\displaystyle y' = (-1)/(1 + \sin x)`

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation