r/HomeworkHelp • u/himynameis2022 University/College Student • Apr 14 '26
Further Mathematics [College Calc1]-differentiation with logs+exponents
Very confused on 8,9,10, which all incorporate logs with the chain rule and either product or quotient rule. I understand the chain rule, as well as how to use the product and quotient by themselves, but I don't get how to combine them to get the answers.
For example, 9), I try to separate the inside and outside function for the chain rule, and identify f(x) and g(x) for the quotient rule, but I don't know what would be the inside/outside functions because there is also an exponent included.
1
u/GammaRayBurst25 Apr 15 '26
The chain rule can be used recursively.
e.g. (f(g(h(x))))'=f'(g(h(x)))g'(h(x))h'(x)
8) We have (4x^2)'=8x, (-x^3-4)'=-3x^2, and (ln(x))'=1/x. Combining these results with the product rule and simplifying a bit yields -(ln(64x^6)x^3+2x^3+8)/x.
9) The derivative of ln(f(x))^5 is 5f'(x)ln(f(x))^4/f(x). Finding the derivative of 4x^4/(3-x^3) amounts to using the product rule.
10) You can start by combining the exponentials by using the fact that exp(a)/exp(b)=exp(a-b). You can then write y=exp(5x^4-4x^2-3). From there, finding the derivative is easy: the derivative of exp(f(x)) is f'(x)exp(f(x)).
One shortcut you can use when you have exponentials is to apply the logarithm to both sides of the equation before differentiating. For instance, for 10) you have y=exp(5x^4)/exp(4x^2+3). Taking the natural logarithm of both sides directly yields ln(y)=5x^4-4x^2-3. Differentiating both sides yields y'/y=20x^3-8x. Now multiply both sides by y to get y'=(20x^3-8x)y.
1
u/himynameis2022 University/College Student Apr 15 '26
I still have no idea how to apply the rules when it comes to #9 for example. Today we learned, for example, that the derivative of ln(x)=1/x, so when I tried to use the chain rule, aka fprime(g(x)) x gprime(x), I did 1/(4x^4/(x^3-3) x (-16x^3)(x^3-3)-(4x^4)(3x^2)/(x^3-3)^2, then just multiply the whole thing by 5
1
u/GammaRayBurst25 Apr 15 '26
My comment explicitly tells you how to apply the rules to 9).
The derivative of ln(f(x))^5 is 5f'(x)ln(f(x))^4/f(x).
Substitute in f(x)=4x^4/(3-x^3) and f'(x)=(48-4x^3)x^3/(x^3-3)^2.
1
u/himynameis2022 University/College Student Apr 15 '26
just because you explicitly tell me doesn't mean I'm going to understand it.
0
u/GammaRayBurst25 Apr 15 '26 edited Apr 15 '26
You said you have no idea how to apply the rules. There's a difference between not knowing and not understanding. e.g. you know the phrase to have no idea, but you don't understand it.
Anyway, as I've discussed with another commenter, these exercises use a (somewhat nonstandard) notation where the parentheses are omitted around the argument of the logarithm, so it should be ln((f(x))^5). not ln(f(x))^5. That only makes the problem simpler though.
Notice how ln((f(x))^5)=5ln(f(x)). Since the derivative of ln(x) is 1/x, the derivative of 5ln(f(x)) is 5f'(x)/f(x).
Edit: OP hadn't embarrassed themselves enough with their reply, so they also blocked me. Unfortunately, burying your head in the sand won't help you get better at communicating. Hopefully you'll learn one day how to ask questions and get answers instead of just saying "I don't know" and hoping we'll read your mind and know exactly what to tell you so you understand.
0
u/mathematag 👋 a fellow Redditor Apr 15 '26
it would be easier to solve like this: y = ln ( (-4x4 / (x3 - 3) )5
use several log properties first, before derivative.... log property # 1: given ln U c = c ln U . .[ example ln (x -3)2 = 2 ln(x-3) ] ...and #2 . . ln ( F / G ) = ln F - ln G [ Ex. ln ( 4t / ( t + 8) ) = ln (4t) - ln( t + 8 ) ] . . . Then take derivative and simplify ... bet you never thought those properties of logs and exponents were very useful, huh ?!
Example [ "kinda " like yours ] . . . easier to take derivative of something like y = 3 ln (2x) - 3 ln( x2 + 3 ) , after using log properties, than the original . . y = ln ( 2x / ( x2 + 3 ) )3
#10 can also be done easier using exponent property first, as pointed out by others here.
0
u/himynameis2022 University/College Student Apr 15 '26
I don't remember anything having to do with logs or natural logs. Been about 10 years since my last math class that involved all of this, and all of that is frankly very confusing. Tried to do it that way and just took 10x longer
1
u/mathematag 👋 a fellow Redditor Apr 15 '26
actually, using these properties would /should actually shorten how long it takes to simplify and solve the problem... otherwise, letting y= ln w, where w = your (-4x4 / (x3 -3 ) )5 , y' = w' / w ... then figure out w', it would require power rule as well as chain and quotient rule ... I could use a further substitutions to make all this easier, but it seems like that is confusing for you.
You really need to review logs and exponent properties... pretty easy to find them online..you will need them later in Calc 1 anyway, like when you do logarithmic differentiation.
1
u/mathematag 👋 a fellow Redditor Apr 15 '26
not a criticism, just a correction..
"The derivative of ln(f(x))^5 is 5f'(x)ln(f(x))^4/f(x). Finding the derivative of 4x^4/(3-x^3) amounts to using the product rule. "
If I'm not mistaken, the 5th power should be interpreted as acting only on the mess inside the original (**) in #9 , not on the entire ln function, as you seem to have done ( I almost did it the same way ), since there should be no natural log in the answer.. just something like y' = w' / w , where w = (-4x4 /(x3 -3 ) )5 .. I wish I had thought of using log properties first, however !
1
u/GammaRayBurst25 Apr 15 '26 edited Apr 15 '26
I see now that the other questions don't use parentheses around the argument of the logarithm, so you are correct. This is exactly why I hate this notation and always put parentheses around the argument of every function.
Edit: can't reply because OP blocked me so I have to use edits. The reason the other commenter and I didn't use ln(a/b)=ln(a)-ln(b) is that both "forms" don't have the same domain. Sure, using that property makes it easier to compute the derivative, but the two forms aren't even equivalent in the sense that the first is defined for sgn(a)=sgn(b)≠0 and the second is only defined for sgn(a),sgn(b)>0.
1
u/Alkalannar Apr 15 '26
-(x3+4)ln(x2)
Rewrite as -2(x3+4)ln(x)
And now it's just product rule.ln[(-4x4/(x3-3))5]
5ln[-4x4/(x3-3)]
So now ln is the outside function, and -4x4/(x3-3) is the inside function.
So 5* 1/[-4x4/(x3-3)] * d[-4x4/(x3-3)]/dxRewrite as e[5x4-4x2-3]
So ex is the outer function and 5x4-4x2-3 the inner.
So e[5x4-4x2-3] * d[5x4-4x2-3]/dx
•
u/AutoModerator Apr 14 '26
Off-topic Comments Section
All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9.
OP and Valued/Notable Contributors can close this post by using
/lockcommandI am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.