Calculus II Questions?

Question

Hi, I'm having a problem with the following 2 questions. Am I missing a method of differentiation and a method of solving for x?

1. What is the derivative of f(x)= 4x^(3x) ? I'm assuming the chain rule does not apply here. Because the variable is also in the exponent.

2. f(x)= 5x^3 + 5x - 1 Find the inverse function, let it be g(x), and find g(-11) and g'(-11). For this one, if someone could just help me find the inverse function, I think I'll be alright.

Thanks!

Sorry, for #2, you DON'T need to know how find the inverse function, just g(-11) and g'(-11). which makes a lot of sense now. But I still have one concern..

"f(x)= x^3 + x + 2 = 0"

How can i solve this? Using guess and check it's obviously "x= -1" but is another way to solve this? Thanks.

Answer 1

1)

f(x) = 4x^(3x)

Since you have a function to a function, the best way to solve this is through logarithmic differentiation. Take the ln of both sides.

ln(f(x)) = ln [4x^(3x)]

Note that we can decompose ln(4x^(3x)) as per the log properties.

ln(f(x)) = ln(4) + ln(x^(3x))

Applying the log property once more,

ln(f(x)) = ln(4) + (3x) lnx

Now, we implicitly take the derivative of both sides. Note that the derivative of ln(4) is 0 because it's a constant. Further, when taking the derivative of 3x lnx, we can ignore the 3 while taking the derivative of everything else.

[1/f(x)] [f'(x)] = 3[ln(x) + x(1/x)]

Reducing this,

[1/f(x)] [f'(x)] = 3[ln(x) + 1]

And now, multiplying both sides by f(x), we get

f'(x) = f(x) ( 3 [ln(x) + 1] )

Remember that f(x) = 4x^(3x), so replacing that, we have

f'(x) = 4x^(3x) ( 3 [ln(x) + 1] )

2. f(x) = 5x^3 + 5x - 1

Are you absolutely sure you're asked to find the inverse of this? This is a cubic and is VERY difficult to obtain the inverse.

Answer 2

Use e^ln(4x^3x)

I dont know if you mean (4x)^(3x) or 4 x^(3x).

Use properties of logarythms and the differentiate.

"f(x)= x^3 + x + 2 = 0"

You arent using guess, you are using a criterium for detecting evident roots

You could solve it using the Tartaglia-Cardano method, but its too complicated this way. You can use too the rational roots theorem. The possible rational roots are +/-1 and +/-2.

Ana

Answer 3

The chain rule allways applies but to make it easier if you have the varaiable also in the exponent use this simple trick fom logs

a^b = e ^(loga^b) = e^(b*loga) where the variable is only in the
exponent.

IN your case y=4 e^3x*logx

the derivative is e^3x*log(x) * ( 3logx +3x *1/x )Chain rule)=

=4 x^3x ( 3logx+3)

Answer 4

1. What is the derivative of f(x)= 4x^(3x) ? I'm assuming the chain rule does not apply here. Because the variable is also in the exponent.

f(x) = 4x^(3x)
ln[f(x)] = ln[4x^(3x)] = ln 4 + ln[x^(3x)] = ln 4 + (3x)(ln x)
f'(x)/f(x) = 3ln(x) + 3x/x = 3ln(x) + 3 = 3(ln(x) + 1)
f'(x) = f(x){3(ln(x) + 1)} = 4x^(3x){3(ln(x) + 1)}
= 12x^(3x){(ln(x) + 1)}

Answer 5

1.f(x)=4x^(3x)
use logarithms

lnf(x) =ln(4x^(3x))

lnf(x) =ln4+lnx^(3x)

then differenciate

f('x)/f(x) = d(3xlnx)/dx

f('x)/f(x) =3x(1/x)+3lnx

f('x) = (3+3lnx)f(x)

f('x) =3(1+lnx)4x^(3x)