Let's apply the derivative to both sides with respect to x. We'll use the chain rule.
![y = \ln\left(x + \sqrt{x^2+a^2}\right)\\\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(x + \sqrt{x^2+a^2}\right)\right]\\\\\\\frac{dy}{dx} = \frac{1}{x+\sqrt{x^2+a^2}}*\frac{d}{dx}\left[x + \sqrt{x^2+a^2}\right]\\\\\\\frac{dy}{dx} = \frac{1}{x+(x^2+a^2)^{1/2}}*\left(1 + 2x*\frac{1}{2}(x^2+a^2)^{-1/2}\right)\\\\\\\frac{dy}{dx} = \frac{1}{x+(x^2+a^2)^{1/2}}*\left(1 + x*\frac{1}{(x^2+a^2)^{1/2}}\right)\\\\\\\frac{dy}{dx} = \frac{1}{x+W}*\left(1 + x*\frac{1}{W}\right)\\\\\\](https://tex.z-dn.net/?f=y%20%3D%20%5Cln%5Cleft%28x%20%2B%20%5Csqrt%7Bx%5E2%2Ba%5E2%7D%5Cright%29%5C%5C%5C%5C%5C%5C%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%5Cln%5Cleft%28x%20%2B%20%5Csqrt%7Bx%5E2%2Ba%5E2%7D%5Cright%29%5Cright%5D%5C%5C%5C%5C%5C%5C%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cfrac%7B1%7D%7Bx%2B%5Csqrt%7Bx%5E2%2Ba%5E2%7D%7D%2A%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5Bx%20%2B%20%5Csqrt%7Bx%5E2%2Ba%5E2%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cfrac%7B1%7D%7Bx%2B%28x%5E2%2Ba%5E2%29%5E%7B1%2F2%7D%7D%2A%5Cleft%281%20%2B%202x%2A%5Cfrac%7B1%7D%7B2%7D%28x%5E2%2Ba%5E2%29%5E%7B-1%2F2%7D%5Cright%29%5C%5C%5C%5C%5C%5C%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cfrac%7B1%7D%7Bx%2B%28x%5E2%2Ba%5E2%29%5E%7B1%2F2%7D%7D%2A%5Cleft%281%20%2B%20x%2A%5Cfrac%7B1%7D%7B%28x%5E2%2Ba%5E2%29%5E%7B1%2F2%7D%7D%5Cright%29%5C%5C%5C%5C%5C%5C%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cfrac%7B1%7D%7Bx%2BW%7D%2A%5Cleft%281%20%2B%20x%2A%5Cfrac%7B1%7D%7BW%7D%5Cright%29%5C%5C%5C%5C%5C%5C)
where W = (x^2+a^2)^(1/2) = sqrt(x^2+a^2)
Let's simplify that a bit.
This concludes the first part of what your teacher wants.
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Now onto the second part.
In this case, a = 3 so a^2 = 3^2 = 9.
Recall that if ![g(x) = \frac{d}{dx}\left[f(x)\right]](https://tex.z-dn.net/?f=g%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5Bf%28x%29%5Cright%5D)
then 
We can say that f(x) is the antiderivative of g(x). The C is some constant.
So,
Now let g(x) = ln(x + sqrt(x^2+9) ) + C
Then compute
- g(0) = ln(3)+C
- g(4) = ln(9) = ln(3^2) = 2*ln(3)+C
Therefore,
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Extra info:
Interestingly, WolframAlpha says that the result is 
, but we can rewrite that into ln(3) because inverse hyperbolic sine is defined as
which is the function your teacher gave you, but now a = 1.
If you plugged x = 4/3 into the hyperbolic sine definition above, then you should get 
Sixty two is your answer you’re adding all the sides
Answer:
A. It will be multiplied by 4.
Step-by-step explanation:
When all the dimensions of a 2-dimensional figure are multiplied by a number x then the area is multiplied by x^2. Since all numbers are multiplied by 2 then the area will be multiplied by 2^2 or 4.
To prove this:
12*14=168
Now double the dimensions
24*28=672
Notice that 168*4=672
If this helped please consider giving brainliest.
Answer: B = ( -3 , 2)
Step-by-step explanation:
You would use a mid point formular because you are trying to find the middle of a line . The midpoint formular is (X1 + X2 , Y1 + Y2 )
/2 /2
From the information given A is (7 , -2 )
If you fill it in it would be 7 + X , -2 + X
/2 /2 = (2 , 0 )
The only posible solution would be 7- 3 /2 , -2 + 2 / 2
Makeing B (-3, 2).
