Can somebody please check these over for me? I really want to know what I did wrong if I made mistakes!
Differentiate the following:
1. y = ln [(4-x)/(4+x)]
Since it's the ln of the entire thing, then I can apply ln a - ln b
so, dy/dx = y' = 1/(4-x) *(-1) - 1/(4+x) *(1)
dy/dx = y' = -1/(4-x) - 1/(4+x)
2. y = ln √((3u+2)/(3u-2))
so that's the same as
y = ln ((3u+2)/(3u-2))^(1/2)
= 1/2 ln ((3u+2/3u-2))
using the ln a - ln b property again:
1/2[ln(3u+2) - ln(3u-2)]
dy/dx = 1/2 [1/(3u+2)*3 - 1/(3u-2)*3]
dy/dx = 1/2[3/(3u+2) - 3/(3u-2)
3. y = f(x) = tan(ln(4x+1))
so first I found the derivative of ln 4x+1
= 1/(4x+1)*4 = 4/(4x+1)
I plugged that number back into the original, making it
y = f(x) = tan(4/(4x+1))
Since the d/dx of tan x = sec^2x,
does it just end up being
dy/dx = sec^2(4/(4x+1)) ?
I am confused on how to do a problem like this:
4. f(t) = ln (2t+1)^3/(3t-1)^4
I can't apply the ln a - ln b property since it's not the ln of the entire thing, so how should I do it?
#1 and #2 are correct
in #3
if y = tan(u)
then y' = sec^2 (u) * du/dx
so y = tan(ln(4x+1))
y' = sec^2 [ln(4x+1)]*4/(4x+1)
#4 is a messy quotient rule problem
perhaps you could use the product rule by rewriting it as
f(t) = [3ln (2t+1)][(3t-1)^-4]
It would still be messy to simplify.
To differentiate the function \(f(t) = \ln\left(\frac{{(2t+1)^3}}{{(3t-1)^4}}\right)\), you can use two rules: the chain rule and the power rule. Here's how you can do it step by step:
1. First, apply the power rule to the numerator and the denominator:
\[\frac{{d}}{{dt}}\left((2t+1)^3\right) = 3(2t+1)^2 \quad \text{and} \quad \frac{{d}}{{dt}}\left((3t-1)^4\right) = 4(3t-1)^3.\]
2. Next, use the chain rule to differentiate the entire function. The chain rule states that if \(f(x)\) is a composite function \(g(h(x))\), then its derivative is given by \(f'(x) = g'(h(x))h'(x)\). In this case, the outer function is \(\ln(x)\) and the inner function is \(\frac{{(2t+1)^3}}{{(3t-1)^4}}\).
a. Differentiate the outer function: \(\frac{{d}}{{dt}}\left(\ln(x)\right) = \frac{{1}}{{x}}\).
b. Differentiate the inner function (using the results from step 1):
\(\frac{{d}}{{dt}}\left(\frac{{(2t+1)^3}}{{(3t-1)^4}}\right) = \frac{{3(2t+1)^2}}{{(3t-1)^4}} - \frac{{4(3t-1)^3}}{{(2t+1)^3}}.\)
3. Multiply the results from step 2a and 2b to get the derivative of the entire function:
\[\frac{{d}}{{dt}}\left(\ln\left(\frac{{(2t+1)^3}}{{(3t-1)^4}}\right)\right) = \frac{{1}}{{(2t+1)^3/(3t-1)^4}} \cdot \left(\frac{{3(2t+1)^2}}{{(3t-1)^4}} - \frac{{4(3t-1)^3}}{{(2t+1)^3}}\right).\]
Simplifying this expression further is possible, but the final result above is the derivative of the original function.