Check out some similar questions!

iammario08 · Apr 22, 2009, 08:00 PM

Hello everyone! So my professor gave us 100 derivative problems for our review guide, and I am not sure how to do these ones, please help!

For these you have to find the derivative:

f(x) = ln(9 x+2)

h(x) = e^(x+38)

g(x) = 18^x

g(x) = (5 x^2 -7x)ln(x)

g(x) = (4 x- 6)^(0.5) ln(x)

h(x) = ln[(3x+1)/(4x-4)]

s(x) = ln(7 x - 5)^1.3

s(x) = log_3(9 x^2+4 x)

r(x) = [ln(38 x^2)]^2

r(x) = ln(x+4)+9 x^3 e^x

r(x) = e^(6 x+5)

r(x) = e^(5 x^2-6 x+1/x)

t(x) = 3^(6 x-2)

Find equation of the straight line tangent to the following curve at the point where x = 1. (Round all coefficients to two decimal places.) Use graphing technology to check your answer by plotting the given curve together with the tangent line.
y = ln sqrt(5 x^2+6)

Thanks a bunch!

Perito · Apr 23, 2009, 05:38 AM

$\frac {d}{dx}\,e^x\,=\,e^x$

$\frac {d}{dx}\,ln(x)\,=\,\frac 1x$

So, basically all you need is the above and the chain rule.

For the first one, f(x) = ln(9 x+2),

$f^'\,=\,\frac {d}{dx}\,ln(9 x+2)\,=\,\frac {9}{(9x+2)}$

since

$\frac {d}{dx}\,(9x+2)\,=\,\9$

Unknown008 · Apr 23, 2009, 09:26 AM

Huh, Perito, you missed on thing there.

The derivative of ln (9x+2) is 9/(9x+2)

Ok, I'll post the general way to differenciate those types of equations:

For $ae^{bx}$ your derivative will be $ae^{bx}.f'(bx)$ when e is exponential.

For $alnf(x)$ your derivative will be $\frac{af'(x)}{f(x)}$ when ln is any logarithm (ie ln and log)

EDIT::: Corrected the $\frac{a}{f'(x)}$ Sorry, was tired... with that MUN... rahhhh!

Chris-infj · Apr 23, 2009, 10:41 AM

If y = ln [f(x)], then, dy/dx = f'(x)/f(x) where

but if y = log_3 [f(x)] then dy/dx = [ f'(x)/f(x) ] ln(3). Where log_3 is log to base 3.

Perito · Apr 23, 2009, 12:09 PM

Double post? How did I do that?

Perito · Apr 23, 2009, 12:10 PM

Oops. I thought correctly, I just wrote it wrong.

I did have

$\frac {d}{dx}\,ln(x)\,=\,\frac 1x$

;)

galactus · Apr 25, 2009, 04:11 PM

Find equation of the straight line tangent to the following curve at the point where x = 1. (Round all coefficients to two decimal places.) Use graphing technology to check your answer by plotting the given curve together with the tangent line.
y = ln sqrt(5 x^2+6)

Find the derivative of the given function and plug in x=1 to find the slope, m.

Plug in x=1 into the function as it is to find y. You are given x=1.

Plug all those into y=mx+b and solve for b. You're done.

To find the derivative of $y=ln(\sqrt{5x^{2}+6})$ , use the chain rule.

Rewrite as $\frac{1}{2}ln(5x^{2}+6)$ using the log laws.

The derivative of the inside is $10x$

Since the derivative of ln(x) = 1/x, then we have $\frac{1}{5x^{2}+6}$

Multiply them:

$\frac{1}{2} \;\ \cdot \;\ \frac{10x}{5x^{2}+6}=\fbox{\frac{5x}{5x^{2}+6}}$

The worst part is over... finish up?