3. Compute the following derivatives assuming we don't know...

3. Compute the following derivatives assuming we don't know what functions \(f\) and \(g\) are: (a) (2 points) \(f(x) = f(x^2)\) \(g(x) = g(x^3)\) \(\frac{d}{dx} [f(x^2)g(x^3)] = f(x^2)\frac{d}{dx}[g(x^3)] + g(x^3)\frac{d}{dx}[f(x^2)]\) (b) (2 points) \(\frac{d}{dx} f(g(x^2)) = g(x^2) \cdot \frac{d}{dx}(x^2) = 2x \cdot g(x^2)\) \(= f(g(x^2)) \cdot 2x \cdot g(x^2)\) (c) (2 points) \(\frac{d}{dx} f(cos(x)) = \frac{d}{dx} = -sin(x) = f(cos(x)) \cdot (-sin(x))\) (d) (2 points) \(\frac{d}{dx} x^2 + f(x) + ln(x) + 5^x = \frac{d}{dx}[x^2 + f(x) + ln(x) + 5^x] \) \(= 2x + f(x) + \frac{1}{x} + 5^x \cdot ln(5)\) (e) (2 points) \(\frac{d}{dx} sin(f(x)) = [sin(f(x))] = cos(f(x)) \cdot f(x)\)

Transcript

00:01 So in this question, we assume that f and g are both differentiable functions of x with values in the following table.

00:07 First, they say if h of x is equal to x times f of x plus 4 times g of x, then what is h prime of 1? well, first, i need h prime of x.

00:19 So this will require the product rule in that first term.

00:24 So we're going to have our first factor x times the derivative of the second, f prime of x, plus my second factor, f of x, times the derivative of the first factor, 1.

00:38 Plus the derivative of 4g of x would be 4g prime of x.

00:44 And so if i want h prime of 1, that'll be 1 times f prime of 1 is negative 2, plus f of 1 is 5 times 1, plus 4 times g prime of 1 is 6.

01:03 I'm getting negative 2 plus 5 plus 24.

01:09 That looks like 27 is my answer for number 1.

01:15 In number 2, they say suppose h is equal to f over g, and i want h prime of 2.

01:23 Well, here i'm going to use the quotual.

01:27 So i start with my denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, and that's all over my denominator being squared.

01:46 Now i want h prime of 2.

01:49 So let's see what that's going to give us.

01:53 So i'm going to have g of 2.

01:56 Note g of 2 is 0 times f prime of 2, f prime of 2 was also 0 minus f of 2 that's 4 times g prime of 2 g prime of 2 is 0 and this is over g of 2 being squared so this is actually 0 squared this actually is undefined okay so this is actually undefined okay notice, g of 2 was equal to 0.

02:42 So this function, h of 2 was undefined.

02:47 This should be undefined...

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