Derive: w = Square root of (x^2 · 5^x )3 : So this is what I did: ((x^2 + 5^x)^3)^1/2 = (x^2+5^x) = 3/2(x^2 + 5^x)^1/2 (2x* 5^x + ln(5))
BOB ( Bac of the book) has a different answer, here is my question, do I have to take the derivative again of (x^2 + 5^x) ?
back*
Is it because it's a product rule ? I think that's what I was missing.
hey :)
wait is that a dot or+ between the two terms ??
@Issy14
it's a , which two terms?
it's a * times (multiplication)
so if it was a multiplication why did u write it as addition ?
\[w=\sqrt{(x^{2}*5^x)^3}\]
that's what it is
that's the original.
\[\large \sqrt{(x^2 · 5^x )^3}\] Is this your question?
sorry, wrong sign all the way
ok first u have to write it like this \[(x ^{2}.5^{x})^{\frac{ 3 }{ 2}}\]
Is there \(\cdot\) or \(+\) ??
so this is what i came up so far with: 3/2(x^2* 5^x) (2x)(5^x) + (5^x * ln(5) (x^2)
yup u wrote it as addition in the second step then u derived it as addition and that's why ur answer is wrong
I truly apologize for the mishap
You have to subtract also in the power, 3/2 you are carrying forward, and then you will do (3/2) - 1 in the power..
I did that, at the beginning: I ended up with 1/2 so 3(x^2*5^x)^1/2 and I proceeded from there
Wait, please write your final answer or the step where you have reached till now..
3/2(x^2* 5^x) (2x)(5^x) + (5^x * ln(5) (x^2)
Why you are not taking exponent there???
exponent there in 3/2(x^2* 5^x)^1/2 (2x)(5^x) + (5^x * ln(5) (x^2)
Yep...
And please be careful with brackets.. At last, you have applied product rule which contains two terms, so first term upto ..........^1/2, after this brackets will start...
got it, but then the back of the book has a different answer and I don't know how they got there
\[\large \color{green}{\implies \frac{3}{2}(x^2* 5^x)^{\frac{1}{2}} [(2x)(5^x) + (5^x * \ln(5) (x^2)]}\]
Sorry, in excitement, I wrote somewhat big there...!!! :)
Yes, that's what I have so far, that's where I stopped. I don't know how to proceed from there.
hahahahhaha it's ok, I like math too, it makes think and be patient.
Now, tell me what your BOB, I mean what your back of book says.. Please write that answer too..
Just for the record, are you SURE that "derive" is a imperative form synonymous with "find the derivative"?
ok, one second BOB says: 3/2 x^2 square root of 5^2x (2 + x ln(5))
If they factored, idk where they got the common factor from
Is that 5^(2x) there??
yes =(
And you should be happy that you have done absolutely right...
so it's BOB crazy?
nuuuuu, don't say that, sometimes BOB has been known to be wrong
Oh sorry, you must be happy now that we are absolutely and perfectly right... Sorry, I did not read the BOB answer carefully... Yes we are right, just we have to rearrange our reached answer..
hahhahahaha...
See, this you know: \[\large (a \cdot b)^x = a^x \cdot b^x\]
And now just take \(x \cdot 5^x\) common from the brackets... And see what you get..
And you must be knowing: \[\large (a)^{\frac{1}{2}} = \sqrt{a}\]
5x? shouldn't it be 5^x that's what you meant right?
\[\large (x^2 \cdot 5^x)^{\frac{1}{2}} = (x^2)^{\frac{1}{2}} \cdot (5^x)^{\frac{1}{2}}\]
ok, one second let's see.
Where???
Ha ha ha ha... :P
Where you have seen 5x ??? Open your eyes, there is no 5x anywhere... :)
omg, I think I finally lost it.
See do step by step..
BOB was right once again... it's a hate-love relationship. 3/2 x^2 square root 5^x ( 2 + ln(5))
You got it or not??
Yeah I got it. I wrote it on my whiteboard
Okay, that's nice.. :) Good Luck...!!!
And Well Done!!!
Thank you!!! my brain is sore like a muscle, I'm going to drink water.
Okay!!
Except that \(\sqrt{x^{2}} = |x|\) in the absence of additional information.
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