1. Find dy/dx for each of the following. (You do not need to simplify your answers)

a) y = loge 3x + 8x³

b) y = e^x cosec x

c) y = e^4x+3/4x + 3

d) y = ³√2x - 1

2. If g(x) = 7x³ - 2x² + 4x - 8 find the third derivative g'''(x)

Printable View

- Apr 10th 2006, 03:02 PMKiwigirlDifferentiate functions
1. Find dy/dx for each of the following. (You do not need to simplify your answers)

a) y = loge 3x + 8x³

b) y = e^x cosec x

c) y = e^4x+3/4x + 3

d) y = ³√2x - 1

2. If g(x) = 7x³ - 2x² + 4x - 8 find the third derivative g'''(x) - Apr 10th 2006, 11:35 PMearbothQuote:

Originally Posted by**Kiwigirl**

with most of your problems you have to use the chain rule. With b) I used the quotient rule and with c) you have to controle if I wrote the function correctly:

to a):

$\displaystyle y=\ln(3x)+8x^3 \Longrightarrow y'=\frac{1}{3x} \cdot 3+3 \cdot 8x^2$

to b):

$\displaystyle y=\frac{e^x}{\cos(x)} \Longrightarrow y'=\frac{e^x*\cos(x)-e^x*(-\sin(x))}{(\cos(x))^2}$

to c):

$\displaystyle y=e^{4x}+\frac{3}{4x}+3 \Longrightarrow y'=e^{4x} \cdot 4+\frac{3}{4} \cdot \left(-\frac{1}{x^2} \right)$

to d):

$\displaystyle y=\sqrt[3]{2x}-1 \Longrightarrow y'=\frac{1}{3} \cdot (2x)^{-\frac{2}{3}} \cdot 2$

To 2.: You have a function of grade 3. You want to get the third derivative. That must be a constant. So g'''(x)=3!*7=42.

Greetings

EB - Apr 12th 2006, 01:31 PMKiwigirl
Thanks earboth for your quick response :) I really appreciate it!