what is the derivative of e^(3*ln(x^2))

i keep getting 6/x * e^(3*ln(x^2))

but that's not one of the multiple choice so am i doing something wront.

also, what is the limit of (1-cosx)/(2*(sinx)^2) as x approaches zero. is the answer zero or nonexistent

The derivative of e^(3*ln(x^2)):
Let's write ln (x^2) as 2 ln x. Then
d/dx [e^(6 ln x)] = [e^(6 ln x)] * (6/x)

The limit of (1-cosx)/(2*(sinx)^2) as x approaches zero. The answer is NOT zero or undetermined.

Use L'Hopital's rule. The ratio of derivatives of numerator and denominator is
sin x/[4 sin x *cos x] = 1/(4 cos x)
At x = 0, that becomes 1/4

To find the derivative of e^(3*ln(x^2)), you can use the chain rule. First, rewrite ln(x^2) as 2ln(x) since the logarithm of a power can be written as the product of the logarithm of the base and the exponent.

Now, let's call e^(3*ln(x^2)) as y, so y = e^(3*2ln(x)) = e^(6ln(x)) = x^6.

To find the derivative, take the natural logarithm of both sides: ln(y) = ln(x^6).

Now, differentiate implicitly with respect to x: (1/y) * dy/dx = (1/6) * (6/x).

Simplify: dy/dx = (1/y) * (1/6) * (6/x) = (1/6) * (1/x) * (x^6) = x^5/6.

So, the derivative of e^(3*ln(x^2)) is x^5/6.

As for the second question, you need to find the limit of (1-cosx)/(2*(sinx)^2) as x approaches zero.

Substituting x = 0 directly gives us an indeterminate form of 0/0.

To evaluate this limit, we can use L'Hopital's rule.

First, take the derivative of the numerator and denominator separately:

The derivative of 1-cosx is sinx.
The derivative of 2*(sinx)^2 is 4sinx*cosx.

Now, evaluate the limit again using the derivatives:

lim(x→0) (sinx)/(4sinx*cosx).

Since sinx appears in both the numerator and denominator, we can cancel it out:

lim(x→0) 1/(4cosx).

Now, substitute x = 0 into the expression: 1/(4cos(0)) = 1/4.

Therefore, the limit of (1-cosx)/(2*(sinx)^2) as x approaches zero is 1/4.