1. Find (dy)/(dx) where, y = ln((-1)/((x-1))) a....

1. Find $\frac{dy}{dx}$ where, $y = ln(\frac{-1}{(x-1)})$ a. $\frac{1}{(x-1)}$ b. $\frac{-1}{(x-1)}$ c. $\frac{1}{(x-1)^2}$ d. $\frac{-1}{(x-1)^2}$ e. none 2. Find an equation for the tangent line to the graph of $f(x) = 2\sqrt{x}$ at $x = 4$ a. $y - 2 = \frac{1}{2}(x - 2)$ b. $y - 2 = \frac{1}{4}(x - 4)$ c. $y - 4 = \frac{1}{2}(x - 4)$ d. $y - 4 = \frac{1}{4}(x - 4)$ e. none 3. Implicitly, find $\frac{dy}{dx}|(0,0)$ where $y + cos y = x$ a. $\frac{-1}{2}$ b. $\frac{1}{2}$ c. $-1$ d. $1$ e. none 4. If $f(x) = e^{2x+1}$, then $f'(\frac{-1}{2}) = $ a. 0 b. 1 c. 2 d. -1 e. none 5. If $f(x) = x^3 + ln(x)$ then $f''(1) = $ a. 5 b. 7 c. 4 d. 6 e. none 6. If $f(x) = \frac{-(x-1)}{x+1}$ then $f'(x) = $ a. $\frac{1}{(x+1)^2}$ b. $\frac{-1}{(x+1)^2}$ c. $\frac{2}{(x+1)^2}$ d. $\frac{-2}{(x+1)^2}$ e. none 7. If $f(x) = ln(e^x)$ then $f'(x) = $ a. $e^x$ b. $ln(e^x)$ c. 1 d. $\frac{1}{e^x}$ e. none

Transcript

00:01 Alright, so question one.

00:02 We were given that f of x equals 5 times e to the 4x minus 9x, and we want to find the derivative, f prime.

00:21 Okay, so f of x is equal to 5 times e to the negative 5x.

00:34 So then f prime, we have this multiple of 5, and we're going to multiply that by the derivative of this function here.

00:42 So whenever you have e to the x, you're going to write e to whatever that power is.

00:48 And then multiply by the derivative of the exponent by the chain rule.

00:54 So the derivative of negative 5x is just negative 5.

00:58 So then this answer is just negative 25e to the negative 5x.

01:05 For part 2, we are given y equals negative e to the negative 4x over 4.

01:13 And we want to find dy dx.

01:22 Alright, so for dy dx, again, we can go ahead and say that we factor this negative 1 fourth out and multiply it by the derivative of this e to the negative 4x.

01:37 That's just negative 4 times e to the negative 4x, which is then going to give us positive e to the negative 4x, because negative 4 divided by 4 is just 1.

01:51 For 3, we want to find dy dx, for y equals 5 to the negative 2x plus 7.

02:08 All right, so recall that the derivative of a to the x is equal to the natural log of a times a to the x.

02:23 So using that rule, we know here that a equals 5 for our problem.

02:30 So our derivative, y prime, or dy dx i should say, is equal to the natural log of 5 times 5 to the negative 2x plus 7.

02:40 But then by the chain rule, we have to multiply by the derivative of the exponent, which is negative 2.

02:46 So then this equals negative 2 natural log 5 times 5 to the negative 2x plus 7.

02:54 For part 4, we have f of x equals cosine of sine of x.

03:02 Sorry, excuse me, it's cosine of sine of x.

03:11 We want to find f prime of 1.

03:16 Okay, so for this one, we have sine of x on the inside of cosine, so we're going to use the chain rule.

03:22 So using the chain rule, we take the derivative of the outside, which is cosine.

03:26 So the derivative of cosine is negative sine, so we're going to have negative sine with sine of x on the inside...

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