# Calculus 1 help! Derivative equation.?

I have to calculate the 4th derivative of this equation:

f(x)=2x^4+3sin2x+(2x+1)^4 Then I have to find the value of the 4th derivative at π/12

*π=pi. Just for clarification

Update:

I took the derivative four times. I just wanted to check the answer I got (264) with someone else.

Relevance
• KY
Lv 7
10 years ago

I got 456 as the answer; here's my work step-by-step.....

f(x)=2x^4+3sin2x+(2x+1)^4

f'(x) = 8x^3 + 8(2x+1)^3 + 6cos2x

f''(x) = 24x^2 + 48(2x+1)^2 - 12sin2x

f'''(x) = 48x + 192(2x+1) - 24cos2x

f''''(x) = 48 + 384 + 48sin2x = 432 + 48sin2x

f''''(π/12) = 432 + 48sin{2(π/12)}

f''''(π/12) = 432 + 48sin{π/6}

f''''(π/12) = 432 + 48(1/2)

f''''(π/12) = 432 + 24

f''''(π/12) = 456

• 10 years ago

Take the derivative 4 times

• 10 years ago

456

• 10 years ago

Some of the constants will drop out

• 10 years ago

I got 456 too.

• 10 years ago

Ok I got 113π/3 which is probably wrong. I'll keep working on it.

• 10 years ago

Use a graphing caluator

• 10 years ago

I concur with that earlier post