replied 5 years ago.
Ok, I can definitely help you with the first part.
So, a stationary point is the same as a critical point (if you know that term better). Basically we need to find where the derivative is equal to 0. So, let's differentiate f(x).
Now, set the derivative equal to 0, and you can solve for what the stationary points are.
Therefore, x=-4 and x=1 are your stationary points.
B. To use the first derivative test, we will test what the derivative does to the left of x=-4 and to the right of x=-4.
To the left of x=-4, f'(x)0.
When the slope goes from negative to positive, we have a relative minimum.
For the second derivative test, we must of course first find the the second derivative.
Now, we test f''(1). If it is less than 0, we have a maximum, and if it is greater than 0, we have a minimum.
Therefore, we have a relative maximum.
We can also verify this using the first derivative test, since to the left of x=1 (and to the right of x=-4), f'(x)>0 and to the right of x=1, f'(x)<0, which means we have a relative maximum.
g'(x)=5cos(5x) (this involves the chain rule where we must take the derivative of the outer function (sin) and then multiply it by the derivative of the inside function (5x) which is 5).
The derivative using the product rule is fg'+gf'.
So, we take x^4(5cos(5x))+sin(5x)(4x^3)
Now for the second one:
f'(t)=6e^6t, since we take the derivative of e^t which is just e^t and multiply it by the derivative of 6t (which is 6). This is basically u substitution where u=6t and du=6.
The quotient rule is (gf'-fg')/g^2 PLEASE NOTE: In the quotient rule the top (gf'-fg') matters in terms of order. The product rule does not. It can be written as fg'+gf' or gf'+fg'.
So, we take ((t^3+8)(6e^6t)-(e^6t)(3t^2))/(t^3+8)^2
Now to the next part,
This utilizes the chain rule again.
We take the derivative of the outside, 6cos, which is -6sin (please don't forget that cosine has the negative when you take the derivative, whereas the derivative of sine does not). Then take the derivative of the inside ((1/3)x) which is 1/3.
We get, -6sin((1/3)x)*1/3, which is -2sin((1/3)x).
So, using the composite rule (AKA chain rule), we basically take the derivative of the outside, ln(u) (where u is the inside function), which is 1/u du.
du is the derivative of the inside, namely, -2sin((1/3)x).
So, the answer is 1/(6cos(1/3x))*(-2sin((1/3)x)).
B. (First of all, the instructions must be wrong here, since an indefinite integral has no bounds. Definite integration involves the bounds from 6 to 8).
So, this involves u substitution. Make u the function that will result in giving you what is leftover with the dx.
So, let's try u=4+7x^2 since that will leave us with something times xdx. du is the derivative of u. So, du=14x, which means du/14=xdx.
So, we now have (1/14)int(u du,6,8). We put the constant that goes with du in front of the integral when we integrate. The integral of u, is u^2/2, which when we replace is (4+7x^2)/2 from 6 to 8.
Start with the upper bound (8), that gives us, 226. Now plug in the lower bound (6), which gives us 128. Now subtract the lower bound's value from the upper bound's value. 226-128=98.
Therefore the answer is 98.
That's what I can help you with at this point. I really hope this is helpful, and if anything doesn't make sense, please feel free to ask me. I hope another expert can help you with the rest of the problems on here. I'll look some things up though, and see if I can come up with some things for your other problems.