Derivatives: Learning Objectives 6, 7
1. Find $F'(x)$ if $F(x) = \sqrt{x^2 + 1}$.
Identify the outer and inner functions and apply the chain rule. Let the outer function be $f(u) = \sqrt{u}$ and the inner function be $g(x) = x^2 + 1$.
$F'(x) = \frac{x}{\sqrt{x^2 + 1}}$
Let $f(u) = \sqrt{u} = u^{1/2}$ and $g(x) = x^2 + 1$.
The derivatives are $f'(u) = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}}$ and $g'(x) = 2x$.
By the chain rule, $F'(x) = f'(g(x)) \cdot g'(x)$.
$$F'(x) = \frac{1}{2\sqrt{x^2 + 1}} \cdot (2x) = \frac{x}{\sqrt{x^2 + 1}}$$
2. Differentiate $y = (x^3 - 1)^{100}$.
This is a direct application of the chain rule, sometimes called the general power rule.
$y' = 300x^2(x^3 - 1)^{99}$
Let the outer function be $f(u) = u^{100}$ and the inner function be $g(x) = x^3 - 1$.
Their derivatives are $f'(u) = 100u^{99}$ and $g'(x) = 3x^2$.
By the chain rule, $y' = f'(g(x)) \cdot g'(x)$.
$$y' = 100(x^3 - 1)^{99} \cdot (3x^2) = 300x^2(x^3 - 1)^{99}$$
3. Find $g'(x)$ if $g(x) = (2x + 1)^5(x^3 - x + 1)^4$.
Start with the product rule, then use the chain rule to differentiate each of the two factors.
$g'(x) = 10(2x + 1)^4(x^3 - x + 1)^4 + 4(3x^2 - 1)(2x + 1)^5(x^3 - x + 1)^3$
Let $u(x) = (2x + 1)^5$ and $v(x) = (x^3 - x + 1)^4$. By the product rule, $g'(x) = u'(x)v(x) + u(x)v'(x)$.
First, find $u'(x)$ using the chain rule: $$u'(x) = 5(2x + 1)^4 \cdot \frac{d}{dx}(2x+1) = 5(2x + 1)^4 \cdot 2 = 10(2x + 1)^4$$ Next, find $v'(x)$ using the chain rule: $$v'(x) = 4(x^3 - x + 1)^3 \cdot \frac{d}{dx}(x^3 - x + 1) = 4(x^3 - x + 1)^3 \cdot (3x^2 - 1)$$ Substitute these into the product rule formula: $$g'(x) = \left[10(2x + 1)^4\right] \left[(x^3 - x + 1)^4\right] + \left[(2x + 1)^5\right] \left[4(x^3 - x + 1)^3(3x^2 - 1)\right]$$ $$g'(x) = 10(2x + 1)^4(x^3 - x + 1)^4 + 4(3x^2 - 1)(2x + 1)^5(x^3 - x + 1)^3$$
First, find $u'(x)$ using the chain rule: $$u'(x) = 5(2x + 1)^4 \cdot \frac{d}{dx}(2x+1) = 5(2x + 1)^4 \cdot 2 = 10(2x + 1)^4$$ Next, find $v'(x)$ using the chain rule: $$v'(x) = 4(x^3 - x + 1)^3 \cdot \frac{d}{dx}(x^3 - x + 1) = 4(x^3 - x + 1)^3 \cdot (3x^2 - 1)$$ Substitute these into the product rule formula: $$g'(x) = \left[10(2x + 1)^4\right] \left[(x^3 - x + 1)^4\right] + \left[(2x + 1)^5\right] \left[4(x^3 - x + 1)^3(3x^2 - 1)\right]$$ $$g'(x) = 10(2x + 1)^4(x^3 - x + 1)^4 + 4(3x^2 - 1)(2x + 1)^5(x^3 - x + 1)^3$$
4. Find the derivative of $f(x) = \ln(x^2 + 1)$.
Use the chain rule with the outer function being $\ln(u)$ and the inner function being $u = x^2 + 1$.
$f'(x) = \frac{2x}{x^2 + 1}$
Let the outer function be $h(u) = \ln(u)$ and the inner function be $g(x) = x^2 + 1$.
The derivatives are $h'(u) = \frac{1}{u}$ and $g'(x) = 2x$.
By the chain rule, $f'(x) = h'(g(x)) \cdot g'(x)$.
$$f'(x) = \frac{1}{x^2 + 1} \cdot (2x) = \frac{2x}{x^2 + 1}$$
5. Given that $\frac{d}{dx}e^x = e^x$, find the derivative of the general exponential function, $b^x$.
Use the identity $b = e^{\ln(b)}$ to rewrite the function $b^x$. Then apply the chain rule. This is a standard derivation.
$\frac{d}{dx}b^x = b^x \ln(b)$
We rewrite $b^x$ using the identity $b = e^{\ln b}$.
$$b^x = (e^{\ln b})^x = e^{x \ln b}$$
Now we differentiate $e^{x \ln b}$ with respect to $x$. This requires the chain rule. Let the outer function be $f(u)=e^u$ and the inner function be $u(x) = x \ln b$.
The derivative of the inner function is $\frac{du}{dx} = \ln b$ (since $\ln b$ is a constant).
Applying the chain rule:
$$\frac{d}{dx} e^{u} = e^u \cdot \frac{du}{dx}$$
$$\frac{d}{dx} e^{x \ln b} = e^{x \ln b} \cdot (\ln b)$$
Since $e^{x \ln b} = b^x$, we have:
$$\frac{d}{dx}b^x = b^x \ln(b)$$
6. Find $y'$ if $y = x^2e^{-3x}$.
Apply the product rule. The derivative of the exponential term will require the chain rule.
$y' = 2xe^{-3x} - 3x^2e^{-3x}$ or $y' = xe^{-3x}(2 - 3x)$
Let $u(x) = x^2$ and $v(x) = e^{-3x}$. By the product rule, $y' = u'v + uv'$.
The derivatives of the parts are: $$u'(x) = 2x$$ $$v'(x) = e^{-3x} \cdot \frac{d}{dx}(-3x) = e^{-3x} \cdot (-3) = -3e^{-3x}$$ Substitute these into the product rule: $$y' = (2x)(e^{-3x}) + (x^2)(-3e^{-3x})$$ $$y' = 2xe^{-3x} - 3x^2e^{-3x}$$
The derivatives of the parts are: $$u'(x) = 2x$$ $$v'(x) = e^{-3x} \cdot \frac{d}{dx}(-3x) = e^{-3x} \cdot (-3) = -3e^{-3x}$$ Substitute these into the product rule: $$y' = (2x)(e^{-3x}) + (x^2)(-3e^{-3x})$$ $$y' = 2xe^{-3x} - 3x^2e^{-3x}$$
7. If $F(x) = e^{-e^{-(x-\theta)}}$, find $f(x) = F'(x)$.
This is a nested chain rule problem involving three functions. Identify the outermost, middle, and innermost functions and apply the chain rule sequentially from the outside in.
$f(x) = F'(x) = e^{-(x-\theta)} e^{-e^{-(x-\theta)}}$
This problem requires applying the chain rule multiple times. Let's define the nested functions:
Substitute the functions back in: $$ F'(x) = e^{h(k(x))} \cdot (-e^{k(x)}) \cdot (-1) $$ $$ F'(x) = e^{-e^{-(x-\theta)}} \cdot (-e^{-(x-\theta)}) \cdot (-1) $$ The two negative signs cancel out, leaving: $$ F'(x) = e^{-e^{-(x-\theta)}} \cdot e^{-(x-\theta)} $$ So, the final derivative is $f(x) = e^{-(x-\theta)} e^{-e^{-(x-\theta)}}$.
- Outermost function: $g(u) = e^u$
- Middle function: $h(v) = -e^v$
- Innermost function: $k(x) = -(x-\theta)$
- $g'(u) = e^u$
- $h'(v) = -e^v$
- $k'(x) = -1$
Substitute the functions back in: $$ F'(x) = e^{h(k(x))} \cdot (-e^{k(x)}) \cdot (-1) $$ $$ F'(x) = e^{-e^{-(x-\theta)}} \cdot (-e^{-(x-\theta)}) \cdot (-1) $$ The two negative signs cancel out, leaving: $$ F'(x) = e^{-e^{-(x-\theta)}} \cdot e^{-(x-\theta)} $$ So, the final derivative is $f(x) = e^{-(x-\theta)} e^{-e^{-(x-\theta)}}$.
8. Let $\ell(\lambda) = n\log(\lambda) - \lambda \sum_{i=1}^n x_i$. Find $\frac{d\ell}{d\lambda}$. You may assume $x_1...x_n$ are known.
Differentiate with respect to $\lambda$. Treat $n$ and the summation term $\sum x_i$ as constants.
$\frac{d\ell}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^n x_i$
We differentiate $\ell(\lambda)$ term by term with respect to $\lambda$.
For the first term: $$\frac{d}{d\lambda}(n\log(\lambda)) = n \cdot \frac{1}{\lambda} = \frac{n}{\lambda}$$ For the second term, let the constant $C = \sum_{i=1}^n x_i$. $$\frac{d}{d\lambda}(-\lambda C) = -C = -\sum_{i=1}^n x_i$$ Combining the results gives: $$\frac{d\ell}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^n x_i$$
For the first term: $$\frac{d}{d\lambda}(n\log(\lambda)) = n \cdot \frac{1}{\lambda} = \frac{n}{\lambda}$$ For the second term, let the constant $C = \sum_{i=1}^n x_i$. $$\frac{d}{d\lambda}(-\lambda C) = -C = -\sum_{i=1}^n x_i$$ Combining the results gives: $$\frac{d\ell}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^n x_i$$
9. If $p(\beta_1) = \frac{1}{1 + e^{-(\beta_0 + \beta_1 x)}}$, find $\frac{dp}{d\beta_1}$. You may assume $x$ and $\beta_0$ are known.
Differentiate with respect to $\beta_1$, treating $x$ and $\beta_0$ as constants. Rewrite the expression as $(1 + e^{-(\beta_0 + \beta_1 x)})^{-1}$ and apply the chain rule multiple times.
$\frac{dp}{d\beta_1} = \frac{x e^{-(\beta_0 + \beta_1 x)}}{(1 + e^{-(\beta_0 + \beta_1 x)})^2}$
Rewrite $p$ as $p = (1 + e^{-(\beta_0 + \beta_1 x)})^{-1}$. We apply the chain rule. Let $u = 1 + e^{-(\beta_0 + \beta_1 x)}$. Then $p = u^{-1}$.
$$\frac{dp}{d\beta_1} = \frac{dp}{du} \frac{du}{d\beta_1} = -u^{-2} \cdot \frac{du}{d\beta_1} = \frac{-1}{(1 + e^{-(\beta_0 + \beta_1 x)})^2} \cdot \frac{du}{d\beta_1}$$
Now we find $\frac{du}{d\beta_1}$:
$$\frac{du}{d\beta_1} = \frac{d}{d\beta_1} (1 + e^{-(\beta_0 + \beta_1 x)}) = \frac{d}{d\beta_1} (e^{-(\beta_0 + \beta_1 x)})$$
We use the chain rule again. Let $v = -(\beta_0 + \beta_1 x)$. Then $\frac{dv}{d\beta_1} = -x$.
$$\frac{d}{d\beta_1} e^v = e^v \cdot \frac{dv}{d\beta_1} = e^{-(\beta_0 + \beta_1 x)} \cdot (-x)$$
So, $\frac{du}{d\beta_1} = -x e^{-(\beta_0 + \beta_1 x)}$.
Substituting this back:
$$\frac{dp}{d\beta_1} = \frac{-1}{(1 + e^{-(\beta_0 + \beta_1 x)})^2} \cdot \left(-x e^{-(\beta_0 + \beta_1 x)}\right) = \frac{x e^{-(\beta_0 + \beta_1 x)}}{(1 + e^{-(\beta_0 + \beta_1 x)})^2}$$
10. If $\ell(k) = n\log(k) + n\log(\lambda) + (k-1)\sum_{i=1}^n\log(x_i) - \sum_{i=1}^n(\lambda x_i)^k$, find $\frac{d\ell}{dk}$. You may assume $x_1...x_n$, $\lambda$, and $n$ are known.
Differentiate term-by-term with respect to $k$, treating all other variables ($\lambda$, $n$, $x_i$) as constants. For the last term, use the derivative rule for $a^x$, which is $a^x \ln(a)$. In this case, the variable is $k$ and the base is $a = \lambda x_i$.
$\frac{d\ell}{dk} = \frac{n}{k} + \sum_{i=1}^n \log(x_i) - \sum_{i=1}^n (\lambda x_i)^k \log(\lambda x_i)$
We differentiate $\ell(\lambda, k)$ term by term with respect to $k$.
1. $\frac{d}{dk}(n\log(k)) = \frac{n}{k}$
2. $\frac{d}{dk}(n\log(\lambda)) = 0$ (constant w.r.t. $k$)
3. $\frac{d}{dk}\left((k-1)\sum_{i=1}^n\log(x_i)\right) = \frac{d}{dk}\left(k\sum_{i=1}^n\log(x_i) - \sum_{i=1}^n\log(x_i)\right)$. Since $\sum\log(x_i)$ is a constant, the derivative is $\sum_{i=1}^n\log(x_i)$.
4. $\frac{d}{dk}\left(-\sum_{i=1}^n(\lambda x_i)^k\right) = -\sum_{i=1}^n\frac{d}{dk}((\lambda x_i)^k)$. Here, the base $a = \lambda x_i$ is a constant. The derivative of $a^k$ w.r.t. $k$ is $a^k \ln(a)$. So, $\frac{d}{dk}((\lambda x_i)^k) = (\lambda x_i)^k \log(\lambda x_i)$.
Summing the derivatives: $$\frac{d\ell}{dk} = \frac{n}{k} + \sum_{i=1}^n \log(x_i) - \sum_{i=1}^n (\lambda x_i)^k \log(\lambda x_i)$$
1. $\frac{d}{dk}(n\log(k)) = \frac{n}{k}$
2. $\frac{d}{dk}(n\log(\lambda)) = 0$ (constant w.r.t. $k$)
3. $\frac{d}{dk}\left((k-1)\sum_{i=1}^n\log(x_i)\right) = \frac{d}{dk}\left(k\sum_{i=1}^n\log(x_i) - \sum_{i=1}^n\log(x_i)\right)$. Since $\sum\log(x_i)$ is a constant, the derivative is $\sum_{i=1}^n\log(x_i)$.
4. $\frac{d}{dk}\left(-\sum_{i=1}^n(\lambda x_i)^k\right) = -\sum_{i=1}^n\frac{d}{dk}((\lambda x_i)^k)$. Here, the base $a = \lambda x_i$ is a constant. The derivative of $a^k$ w.r.t. $k$ is $a^k \ln(a)$. So, $\frac{d}{dk}((\lambda x_i)^k) = (\lambda x_i)^k \log(\lambda x_i)$.
Summing the derivatives: $$\frac{d\ell}{dk} = \frac{n}{k} + \sum_{i=1}^n \log(x_i) - \sum_{i=1}^n (\lambda x_i)^k \log(\lambda x_i)$$