Question

find dy/dx for each of the following, simplify too:

Answer 1

1) \[y=5.4x^{5}+6^{\tan x}+\sqrt[5]{x ^{2}+3x}\]

Answer 2

Which term are you stuck on? The middle one? How bout the last one?

The first one should be pretty straight forward, just apply the Power Rule.

Answer 3

i have a bunch of question like this one that i am stuck with and cant answer

Answer 4

\[\large 5.4x^5\]Do you know the derivative of this?

Answer 5

\[27x ^{4}\frac{ dx }{ dy }\]

Answer 6

We're taking our derivative `with respect to x`,
So our derivative operater is \(\large \dfrac{d}{dx}\)

So when you apply the chain rule like you did, you should actually get,\[\large 27x^4\frac{dx}{dx}\]But dx/dx is an insignificant term, so we would simply write this as,\[\large 27x^4\]

Ok good, looks like you understand the power rule.

Answer 7

The next term is an exponential.
Remember how to take the derivative of an exponential?
Here's a quick refresher:\[\large \left(e^x\right)' \qquad=\qquad e^x\]If our base is `any` value besides e, then the rule tells us to multiply by the natural log of the base.
\[\large \left(6^x\right)' \qquad=\qquad 6^x(\ln 6)\]
We could do logarithmic differentiation to show where the log is coming from but I think you should just remember the rule for now.

Answer 8

\[6\sec ^{2}x ?\]

Answer 9

When we differentiate the exponential term, it should give us the same thing back.
The exponential won't change it anyway.
We'll have to multiply it by some stuff though, due to these rules.

Answer 10

Applying the rule for differentiating exponentials gives us this orange part,
and the chain rule gives us this blue part,\[\large \left(6^{\tan x}\right)' \qquad=\qquad 6^{\tan x}\color{orangered}{(\ln6)}\color{royalblue}{\left(\tan x\right)'}\]

Answer 11

The little prime is to show that we still need to take the derivative of it.
And yes you have your derivative for tangent* correct, so let's fill in the pieces.

Answer 12

\[6^{\tan x}(\ln6)(\sec ^{2}x)\] so i guess this is the answer

Answer 13

Yah looks right c:
There are a few rules being applied for that to come about.
You might want to practice a few more exponential terms.

Answer 14

for last part \[(x ^{2}+3x)^{\frac{ 1 }{ 5 }}\]
\[\frac{ 1 }{ 5 } (x ^{2}+3x)^{\frac{ -4 }{ 5 }} (2x+3)\]
?

Answer 15

Yup, good job!

Answer 16

i have 5 more, can we do it here?

Answer 17

Sure, post another one in here. That last one didn't take up much space.

Answer 18

yes, forgot to ask, can we simplify it? this is the tricky part for me sometimes, i ill write the rest in one post

Answer 19

I'm not sure what they mean by simplify...
You can rewrite part of the last term in the denominator by switching the negative exponent.

And then you can combine all the terms into one fraction.

But it's going to look a whole lot worse if we do that...
I'm not sure if that's what they mean by simplify.. hmm

Answer 20

\[1) y=e ^{\csc x} (\ln(x ^{2}+8))\]
\[2) y=\frac{ 3x-4 }{ x ^{2} + 5 }\]
\[3) y=\cos ^{4}((e ^{2x}+11x)^{3})\]
\[4) e ^{y}+x ^{2}y-y ^{2}=x ^{2}y ^{3}-4x ^{3}\]

Answer 21

\[1)e ^{\csc x}((\frac{ 1 }{ x ^{2}+8 })(2x) + (e ^{\csc x}-\csc x \cot x)(\ln(x ^{2}+8))\]

changes my answer, this is what i think is correct, i followed this rule:
\[(uv)\prime = uv \prime + u \prime v \]
u and v here are functions

Answer 22

Look good. Just be careful with the way you write things.
\[\large e^{\csc x}-\csc x \cot x\]

The way you wrote this, it looks like subtraction while it is suppose to be multiplication.

Answer 23

\[2) \frac{ (x ^{2} + 5)(3) - (3x-4)(2x) }{ (x ^{2}+5)^{2} }\]
\[=> \frac{ 3x ^{2} + 15 - 6x ^{2} - 8x }{ x ^{4} + 25}\]
\[==> \frac{ -3x ^{2} -8x+15 }{ x ^{4}+25 }\]
\[===> \frac{ -8x - 3 + 15 }{ x ^{2} + 25 }\]
\[====> \frac{ -8x-3-3 }{ x ^{2} + 5 }\]
\[=====> \frac{ -8x-6 }{ x ^{2} + 5 }\]

how it look?

Answer 24

\[\large \frac{ 3x ^{2} + 15 - 6x ^{2} \color{red}{-} 8x }{x^{4}+25}\]See this red sign?
This should be positive, check the step before this, you forgot to distribute the negative sign.

Answer 25

And the bottom is a problem also.
The square doesn't just square each term.

\[\large (x^2+5)^2 \qquad=\qquad (x^2+5)(x^2+5)\]There is more junk that comes along with squaring.
You could foil this out if you wanted to.
But it's better not to.

Answer 26

\[\frac{ -3x ^{2} + 15 + 8x }{ (x ^{2} + 5 )^{2} }\]

so this is how my final answer should look like? i dont think it can be more simpler than that

only if i factor the numerator, not sure how it will turn out: maybe> \[\frac{ (3x+5)(-x+3) }{ (x ^{2}+5)^{2} }\]

Answer 27

The top doesn't factor.
Those factors don't quite work out the way you think they do D:

The part you posted at the top of your post would be your final answer.

Answer 28

ok, great, you are really helpful, Thank for being patient with me! lets continue \[y=\cos ^{4} ((e ^{2x}+ 11x )^{3})\]
\[=>-4\sin ^{3}((e ^{2x}+ 11x )^{3})(3e ^{2x}+ 11x )^{2}(2e ^{2x} + 11)\]

Answer 29

there is a mistake, 3 should be out of the brackets in the middle part

Answer 30

outside the bracket? Ok great, looks correct :D

Answer 31

can we simplify it?

Answer 32

\[\large -12\sin ^{3}\left[(e ^{2x}+ 11x )^{3}\right](e ^{2x}+ 11x )^{2}(2e ^{2x} + 11)\]You can bring that 3 to the front,
beyond that I don't think you would want to simplify any further.
I guess you could maybe expand out the square and combine terms with the last set of brackets.

Answer 33

See if you expand out that second set of brackets, you don't really get anything nice that you can combine together.
So yah, leave it as a square.
\[\large -12\sin ^{3}\left[(e ^{2x}+ 11x )^{3}\right](e^{4x}+22xe^{2x}+121x^2)(2e^{2x}+11)\]

Answer 34

\[4) e ^{y} + x ^{2}y-y ^{2}=x ^{2}y ^{3}-4x ^{3}\]
\[=>e ^{y}\frac{ dy }{ dx } + (2xy+x ^{2}\frac{ dy }{ dx }) - 2\frac{ dy }{ dx } = (2xy ^{3}+3x ^{2}\frac{ dy }{ dx }^{2})\]\[-12x\]
not sure about this, little bit tricky

Answer 35

I hate Leibniz notation for this type of problem.
So much easier to read with primes.

\[\large e^yy'+2xy+x^2y'-\color{orangered}{2yy'}=2xy^3+\color{orangered}{3x^2y^2y'}-12x\]

Check out these orange terms. You should be getting these values.
Everything else looks good though.

Answer 36

ok, so how can we get y'?

Answer 37

Get everything that has a y' attached to it, and get them all on one side.
Throw everything else to the other side.\[\large e^yy'+x^2y'-2yy'-3x^2y^2y'=2xy^3-12x-2xy\]Then factor y' ouch of each term,\[\large y'(e^y+x^2-2y-3x^2y^2)=2xy^3-12x-2xy\]

Answer 38

Then do some division and boom bam, yer done!

Answer 39

great, ill be back with more question alter if you dont mind, do you want me to post them here:)?

Answer 40

No, post them in a new thread.
I'm all math'd out for now :)

I'll take a look if I can, but there are plenty of people on who can help if I'm not around.

Answer 41

thank you very much