Find the derivatives for each of the following functions. Show your work properly!

Question

Find the derivatives for each of the following functions. Show your work properly!

1. f(x) = 1/(x³ -3x + 7)⁸
2. g(t) = sin (πt⁶) (t + 9)³.
3. h (y) = y cos⁷(2y) + sin(y)/ 3y² - 2

Answer 1

1) f(x) = 1/(x³ - 3x + 7)⁸

Rewrite as f(x) = (x³ -3x + 7)^-8

Apply the power rule and then the chain rule:

f '(x) = -8(x³ - 3x + 7)^-9(3x² - 3)

Rewrite as a single numerator and denominator:

f '(x) = [-8(3x² - 3)]/(x³ - 3x + 7)⁹

2) g(t) = sin (πt⁶) (t + 9)³

Use the product rule (u•v)' = u'v + uv'

where u = sin(πt⁶) and v = (t + 9)³

Calculate u' using the trig rule then the chain rule:

u' = cos(πt⁶)•(6t⁵) = 6t⁵cos(πt⁶)

Calculate v' using the power rule:

v' = 3(t + 9)²

Put them together using (u•v)' = u'v + uv'

g '(t) = 6t⁵cos(πt⁶)•(t + 9)³ + sin(πt⁶)•3(t + 9)²

If you want you can factor out 3(t + 9)²:

g '(t) = 3(t + 9)²[2t⁵(t + 9)cos(πt⁶) + sin(πt⁶)]

3. I'm unsure what the last problem is. It might be

or perhaps:

I'll guess that it's the later:

for ycos⁷(2y) you must use the product rule:

(u•v)' = u'v + uv'

u = y

u' = 1

v = cos⁷(2y) also known as (cos(2y))⁷

v' = 7(cos(2y))⁶•(-sin(2y))•(2) = -14cos⁶(2y)sin(2y)

So the derivative of ycos⁷(2y) is (1)(cos⁷(2y)) + y(-14cos⁶(2y)sin(2y)) = cos⁷(2y) + -14ycos⁶(2y)sin(2y)

for sin(y)/(3y² - 2) use the quotient rule (u/v)' = (u'v - uv')/v²

u = sin(y)

u' = cos(y)

v = 3y² - 2

v' = 6y

So the derivative of sin(y)/(3y² - 2) is [(cos(y)•(3y² - 2) - (sin(y)•(6y)]/(3y² - 2)²

Put these two halves together to get:

h '(y) = cos⁷(2y) + -14ycos⁶(2y)sin(2y) + [(3y² - 2)cos(y) - 6ysin(y)]/(3y² - 2)²