Could someone please check my work on the following problems? I want to be sure that I'm doing them correctly; i'm just required to find the derivatives and then simplify.
g(t)= t/sint
g'(t)=1*sint-cost*t/(sint)^2
g'(t)=sint-cost*t/(sint)^2
y=ln(e^ax)
y'=1/e^ax*e^x*a
and i'm sort of stuck on this one
y=e^2x[sin(3x)^2]^2
Y'=2^ex*sin(3X)^2+e^2x[2cos3x]
i got this far, but i'm not sure how to simplify that.
my version:
g(t)= t/sint
g'(t) = (sin t -t cos t)/sin^2t
your version:
g'(t)=1*sint-cost*t/(sint)^2
g'(t)=sint-cost*t/(sint)^2
y=ln(e^ax)
my version:
y = a x ln (e) = a x
y' = a
your version:
y'=1/e^ax*e^x*a times a because d/dx(e^ax) = a e^ax
but that is just x a
and i'm sort of stuck on this one
y=e^2x[sin(3x)^2]^2
my version:
[sin(3x)^2]^2 * 2 e^2x + e^2x * 2 [sin(3x)^2]^1 * d/dx(sin(3x)^2)
= 2 e^2x *{[sin(3x)^2]^2 + sin(3x)^2 * sin(3x)^2* [cos(3x)^2] *2*3x*3]
= 2 e^2x * sin(3x)^2]^2 *{ 1 + 18xcos(3x)^2}
wow - check my arithmetic
Your version:
Y'=2^ex*sin(3X)^2+e^2x[2cos3x]
To check your work, let's go through each problem and compare your answers with the correct solutions.
1. g(t) = t/sin(t)
To find the derivative of this function, you can use the quotient rule, which states that for a function f(t) = u(t)/v(t), where u(t) and v(t) are both differentiable functions:
f'(t) = (v(t) * u'(t) - u(t) * v'(t)) / [v(t)]^2
Applying the quotient rule to g(t) = t/sin(t):
g'(t) = (sin(t) * 1 - t * cos(t)) / sin^2(t)
= sin(t) - t * cos(t) / sin^2(t)
Your answer, g'(t) = sin(t) - cos(t) * t / sin^2(t), is correct.
2. y = ln(e^ax)
To find the derivative of this function, you can apply the chain rule. The chain rule states that for a function f(g(x)), the derivative is given by:
f'(g(x)) * g'(x)
In this case, f(g) = ln(g) and g(x) = e^ax.
So, f'(g) = 1/g and g'(x) = ae^ax.
Therefore, y' = 1 / (e^ax) * ae^ax = a.
Your answer, y' = 1 / (e^ax) * e^x * a, is not correct. The correct result is simply y' = a.
3. y = e^2x[sin(3x)^2]^2
To simplify this expression, start by applying the power rule. The power rule states that if you have a function f(x)^n, then the derivative is given by:
(f(x)^n)' = n * f(x)^(n-1) * f'(x)
Let's break down the expression step by step:
- First, take the derivative of e^2x, which is simply 2e^2x.
- Next, focus on the inner part, [sin(3x)^2]^2. Apply the power rule:
- (sin(3x)^2)' = 2sin(3x) * cos(3x) (using the chain rule and power rule)
Now, putting it all together:
y' = 2e^2x * [sin(3x)^2]^2 + e^2x * 2sin(3x) * cos(3x)
Your answer, y' = 2^ex * sin(3x)^2 + e^2x * 2cos(3x), is not correct. The correct result is:
y' = 2e^2x * [sin(3x)^2]^2 + e^2x * 2sin(3x) * cos(3x)
Remember, simplifying the expression further may not be possible or necessary in some cases, so leaving it as is would be acceptable.
To summarize:
1. g'(t) = sin(t) - t * cos(t) / sin^2(t) (Correct)
2. y' = a (Correct)
3. y' = 2e^2x * [sin(3x)^2]^2 + e^2x * 2sin(3x) * cos(3x) (Correct)