# Calc

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Can anyone help me find the derivative of these functions?

y = x^2x
y = x^-cos(x)
y = ln(7x^2 + 5y^2)

sqrt(3x+y)=2+x^2y^2
ysec(x) = 3xtan(y)
y = arctan[(sqrt((1 + x)/(1 - x)))]

• Calc -

For powers, the following transformation can be useful:
xn = enlog(x)

For example:
xn = en*ln(x)
The derivative can then be obtained by applying the chain rule:
d(xn)/dx
=en*d(n*ln(x))/dx
=en*(n/x)
=nxn-1

By applying the same transformation, and using the chain rule and the product rule, we get:
d(x2x)/dx
=d(e2x*log(x)
=e2x*log(x)*d(2x*log(x))/dx
=e2x*log(x)(2log(x)+2x/x)
=2x2x*(log(x)+1)

The second one can be found similarly.
The third one can be solved using the transformation:
y=ln(7x²+5y²)
ey=7x²+5y²
Apply implicit differentiation to complete.
The rest are also examples of implicit differentiation. Give it a try and post if you have difficulties.

• Calc -

Thank for replying =D. Unfortunately, I have been using implicit differentiation and I do not fully understand the logarithm. I am still getting an incorrect answer. Please show me an example on how to solve these problems. Thank you.

• Calc -

Basically, implicit differentiation is like the chain rule, except that you will leave as is any dy/dx that you encounter.

After that, you will isolate dy/dx on the left hand side, and whatever is left on the right hand side is your answer. Note that the right-hand-side expression may contain the variable y.

For example, given
y²=4xy+3x
You would apply the chain rule on both sides to get:
2y(dy/dx) = 4y + 4x(dy/dx) + 3
Isolate and collect dy/dx
(2y-4x)(dy/dx) = 4y+3
dy/dx = (4y+3)/(2y-4x)
=(1/2)(4y+3)/(y-2x)

You can do the third to the fifth problem using implicit differentiation, along with other rules and techniques.

If you encounter difficulties, post what you have, and we should be able to help you out.

Keep up the good work!

• Calc -

Thank you! For
y sec(x) = 3x tan(y)
I got dy/dx = (3 tan(y) - y tan(x)sec(x)) / (sec(x) - 3x secĀ²(y))
but it was still incorrect...

I have no idea how to do y = arctan[(sqrt((1 + x)/(1 - x)))]
because i never did this type of problem before, help please.

here is another problem that I do not know what I'm doing wrong:
(a) Find the second derivative y '' by implicit differentiation if:
x3 + y3= 27
Do not solve for y in terms of x.
y'' = -2x(y^3+x^3)/y^5

(b) Now simplify your answer to (a) as much as possible, still without solving for y in terms of x.
y'' = -54x/y^5
(but I got this one correct)

Every other problem, I figured it out. Thank you so much, I really appreciate it.

• Calc -

For y sec(x) = 3x tan(y), I got the same answer as yours.
If you have the answer, can you see if you can transform one to the other. If the answer is not available, as in a computerized exercise, you can ask for an example to see if there is anything that we both did wrong. How many "lives" have you got left?

For the arctan question, it is the same as normal, noting that
d(arctan(x))/dx = 1/(1+x²), and
d(sqrt(x))/dx = (1/2)(1/sqrt(x))
If you proceed, you should get something like:
sqrt(1-x)/(2sqrt(1+x) * (1-x))
or if you rationalize the denominator,
sqrt(1-x²)/(2(1-x²))
Read instructions, sometimes the software does not want you to rationalize the denominator.

For the x3 + y3= 27 problem, I got the same answer as yours in both parts. So it is unlikely that "our" part A answer is incorrect, except for algebraic rearrangements if it is a case of computer answers.

I first got -(2x/y³)(y+x³/y³)
which is equivalent to yours, but it may be accepted if it is computer correction.

Sorry that I could find any mistake in your answers, but at least there is a confirmation.

Good work!

• Calc -

For y sec(x) = 3x tan(y), the direction above it says "Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy." I do now know if that makes any different and I believe I have 4 lives left.

For the x3 + y3= 27 problem, I do not get, so I will ask my instructor about it.

The arctan problem, I solved it thanks to your help. Thank you.

Another problem I have is:
"Prove the formula for (d/dx)(arccos x) by the same method as for (d/dx) (arcsin x)."

Thanks again, I really appreciated your help. =]

• Calc -

For y sec(x) = 3x tan(y), yes it does make a difference if the question wants dx/dy and not dy/dx.
You can redo the implicit differentiation using y as the independent variable and therefore end up calculating dx/dy.
You should get the reciprocal of what you got before.

There are different ways of deriving the derivative of sin-1, but I believe the one expected is using implicit differentiation.

Here's how d(sin-1)/dx can be found:
sin(y) = x
differentiate both sides with respect to x:
cos(y)*dy/dx = 1
dy/dx
= 1/cos(y)
= 1/sqrt(1-sin²(y))
= 1/sqrt(1-x²)

You should be able to do the corresponding derivation for cosine with no difficulties.

• Calc -

Thank you!

• Calc -

You're welcome, it's been a pleasure!

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