How Do You Solve These Calculus Differentiation Problems?

[mcintyre_ie]

Hey
I'd really appreciate some help with the following differntiation problems:

(i) Let f(x) = ax^3 + bx^2 + cx +d, where a, b, c, d are å R.
Given that f(x) has only one local turning point, show that:
b^2 = 3ac
Hence, find the coordinates of the local turning point.

Here are my workings for the first part:
b^2 - 4ac = 0 (from formula)
4b^2 - 4(3a)(c) = 0
4b^2 = 12ac
b^2 = 3ac

For the second part, finding the coordinates, I am not sure what i should do, any advice?

(ii) Using logarithms, or otherwise, differentiate:

y=(e^x^2\sqrt{SinX})/(2x + 1)^3

Ok, so I've been trying to do this, but I'm not sure where the logarithms part comes into it, or where to start, maybe a quotient rule?
Any help is appreciated.

 

[FZ+]

b^2 - 4ac = 0 (from formula)

Whaa? Where did you get this formula from?

I would do it by differentiating:

f'(x) = 3ax^2 + 2bx + c = 0

This gives one turning point, given repeated roots in x.

ie. (2b)^2 - 4*3*a*x = 0

This can be solved easily.

Coordinates? Quadratic formulae, again.
x = -2b/6a (since the square root bit equals 0 for repeated roots)

Substitute, and viola!

y=(e^x^2\sqrt{SinX})/(2x + 1)^3

That does look nasty, doesn't it?
Can you clarify with brackets etc what (e^x^2\sqrt{SinX}) is? Do you mean:

e^((x^2)/(sqrt(sinx)))?

As for logs, perhaps you are meant to log both sides, and differentiate implicitly?

 

[HallsofIvy]

"Here are my workings for the first part:
b^2 - 4ac = 0 (from formula)"

My first reaction was the same as FZ+: "where did you get that formula"? Then I realized that you are using the same letters to mean different things.

Given a cubic ax³+ bx²+ cx+ d, a turning point can only occur where the derivative: 3ax²+ 2bx+ c= 0

That can be solved using the quadratic formula and, in order that there be exactly 1 turning point, there must be exactly one solution so "b²- 4ac= 0" where a,b,c are now the coefficients of ax²+ bx+ c. Putting in the values 3a and 2b for "a","b" gives you the rest:

"4b^2 - 4(3a)(c) = 0
4b^2 = 12ac
b^2 = 3ac"

BAD IDEA! Never use the same letters to represent different things! Or at least tell us you are doing that!

As for finding the turning point: 3ax²+ 2bx+ c= 0 and
b²= 3ac. Okay, then, again, from the quadratic formula,
since the discriminant is 0 x= -2b/(6a). You can plug that into the original cubic to find y (exactly what FZ+ suggested)

Like FZ+, I have no idea what "y=(e^x^2\sqrt{SinX})/(2x + 1)^3" is, in part because of that "\". My guess, like his, is that you meant
y=(e^{x^(x^2/sqrt(sin(x)))})/(2x+1)^3.

log(y)= x^(x^(2/sqrt(sin(x)))))-3log(2x+1) which should be easier to differentiate.

 

[mcintyre_ie]

Ok, thanks for the help with the first part. I should probably have explained that the formula was taken from the quadratic formula x = [-b +/- squrt(b^2-4ac)]/2a, using the b^2-4ac = 0 to show that it has one real root.
I think the reason the question was asked using a,b,c,d was to confuse people trying to do it.

As for the log question, i didnt make a great job of typing it out, so ill try again:
Y = [{e^(x^2)}.sqrt{SinX}]/[(2x + 1)^3

Y = e to the power of x squared by the square root of sinX, divided by (2x + 1) to be cubed.
Thanks again for the help.

 

[mcintyre_ie]

Any ideas on solving the second problem?

 

[himanshu121]

is it

y=\frac{e^{x^2} \sqrt{\sin x}}{(2x+1)^3}

 

[mcintyre_ie]

Yeah, that's it, my latex skills arent really up to scratch yet. Thanks, any clues on solving?

 

[himanshu121]

take log u will get

logy= x^2+\frac{log sinx}{2} - 3log(2x+3)

and now differentiate u will get the desired result