I have to help teach this class tomorrow for my Senior Seminar Math class. The lesson is over Exponential and Logarithmic Functions and L'Hospital's Rule. I have all the problems needed to review this lesson, but i can not solve any of them. My teaching partner and I have had schedule conflicts and we are now on our own. It is late starting but any help or advise would be greatly appreciated. I have not had calculus in 5 years.
1. y=�ã(2)*e^(�ã(2)x) y` =?
2. y=ln(sec^(2)x) y`=?
3. y=9^(2t) y`=?
4. y=[2(x^2+1)]/[�ã(cos 2x)] y`=?
EVALUATE INTEGRALS
5. �ç e^t * cos(3e^t - 2) dt
6. �ç(top: ƒÎ/6, bottom: -ƒÎ/2) (cos t)/(1-sin t) dt
7. �ç [ln(x-5)]/(x-5) dx
8. �ç(top: e, bottom: 1) (1/x) * (1+ 7 ln x)^(-1/3) dx
9. �ç(top:3, bottom: �ã3) dt/(3+t^2)
Solve for Y
10. 3^y = 3 ln x y=?
11. lim (x�¨0) (3^x - 1) / x
12. lim (x�¨0) (4 - 4e^x) / xe^x
13. lim (y�¨0+) e^(-1/y) ln y
My partner has the other 13 problems and his are basically the same. I know this is a lot but any help with any of these would be greatly appreciated.
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I understand that you have a lot of problems related to Exponential and Logarithmic Functions and L'Hospital's Rule that you need help with. I will try my best to guide you through each problem and provide explanations along the way.
Let's start with the first problem:
1. y = √2 * e^(√2x)
To find y', the derivative of y with respect to x, we can use the chain rule. The chain rule states that if we have a function f(g(x)), its derivative is f'(g(x)) * g'(x).
In this case, f(x) = √2 * e^x and g(x) = √2x.
The derivative of f(x) with respect to x is simply e^x since the derivative of e^x is itself.
The derivative of g(x) with respect to x is √2 since the derivative of √2x is √2.
So, applying the chain rule, y' = f'(g(x)) * g'(x) = e^(√2x) * √2.
Moving on to the second problem:
2. y = ln(sec^2x)
To find y', we can use the chain rule once again.
Let u = sec^2x, so y = ln(u).
The derivative of u with respect to x can be found using the chain rule. The derivative of sec^2x is 2sec^2x * tanx.
Now, using the chain rule, y' = (1/u) * du/dx.
Substituting u = sec^2x and du/dx = 2sec^2x * tanx, we have y' = (1/sec^2x) * (2sec^2x * tanx) = 2tanx.
Moving on to the next problem:
3. y = 9^(2t)
To find y', we can simply use the power rule. The power rule states that if we have a function f(x) = kx^n, the derivative of f(x) with respect to x is kf'(x) = knx^(n-1).
In this case, k = 9 and n = 2.
So, y' = 2 * 9^(2t) * ln(9).
Next, let's look at the fourth problem:
4. y = [2(x^2+1)] / √(cos 2x)
To find y', we can apply the quotient rule. The quotient rule states that if we have a function f(x)/g(x), its derivative with respect to x is [g(x)f'(x) - f(x)g'(x)] / [g(x)]^2.
In this case, f(x) = 2(x^2+1) and g(x) = √(cos 2x).
Finding f'(x) and g'(x) is straightforward:
f'(x) = d/dx [2(x^2+1)] = 4x
g'(x) = d/dx [√(cos 2x)] = (-1/2) * √(cos 2x) * sin 2x.
Now, applying the quotient rule, we have y' = [g(x)f'(x) - f(x)g'(x)] / [g(x)]^2 = [(√(cos 2x) * sin 2x) * 2(x^2+1) - 4x * √(cos 2x)] / [√(cos 2x)]^2.
Simplifying further, y' = [(2(x^2+1)sin 2x√(cos 2x) - 4x√(cos 2x)] / cos 2x.
Now, let's move on to the evaluation of integrals section:
5. ∫ e^t * cos(3e^t - 2) dt
This integral can be evaluated using integration by parts.
Let u = cos(3e^t - 2) and dv = e^t dt.
Differentiating u and integrating dv, we have du = (-3sin(3e^t - 2)) * (3e^t) dt and v = e^t.
Applying the formula for integration by parts, the integral becomes:
∫ e^t * cos(3e^t - 2) dt = uv - ∫ v du.
Substituting the values, we have:
= e^t * cos(3e^t - 2) - ∫ e^t * (-3sin(3e^t - 2)) * (3e^t) dt.
Simplifying further, the integral becomes:
= e^t * cos(3e^t - 2) + 9∫ sin(3e^t - 2) * e^(2t) dt.
This integral can be further evaluated using integration by parts or other techniques, depending on the specific problem.
Moving on to the next integral:
6. ∫ (cos t)/(1-sin t) dt
This integral can be evaluated using a trigonometric substitution. Let u = 1 - sin t.
Differentiating u, we get du = -cos t dt.
Substituting the values in the integral, it becomes:
∫ (cos t)/(1-sin t) dt = -∫ du/u.
This can be evaluated as -ln|u| + C.
Finally, we can move on to solving for Y:
10. 3^y = 3 ln x
To solve for y, we can take the logarithm of both sides. Since the logarithm base does not matter, we can use the natural logarithm (ln).
Taking the logarithm of both sides gives:
ln(3^y) = ln(3 ln x)
Applying the power rule of logarithms, we have y * ln(3) = ln(3 ln x).
Simplifying further, we get:
y = ln(3 ln x) / ln(3).
I hope this helps you with your Senior Seminar Math class. Remember to carefully apply the rules and techniques explained for each problem. Good luck with your teaching!