指数函数

I believe that we should compute $\frac{d}{dx} a^x$ (i.e. with respect to x).

To do so, let us define a constant $e$ equal to the limit of $\left(1 + \frac{1}{x}\right)^x$ as $x$ approaches infinity. This limit does exist; however, the proof of its existence requires a long and careful argument involving AM-GM and is beyond the scope of this text.

It can be proven further that $\frac{d}{dx} e^x = e^x.$ Thus, $\frac{d}{dx} a^x = \frac{d}{dx} e^{x\ln a} = \ln a \cdot e^{x\ln a} = \ln a \cdot a^x$, by the chain rule.

Taking the limit definition of the derivative, $$\begin{align*} \frac d{dx}(a^x)&=\lim_{\Delta x\to0}\frac{a^{x+\Delta x}-a^x}{\Delta x}\\\ &=\lim_{\Delta x\to0}\frac{a^x(a^{\Delta x}-1)}{\Delta x}\\\ &=a^x\lim_{\Delta x\to0}\frac{a^{\Delta x}-1}{\Delta x}=a^x\lim_{\Delta x\to0}\frac{a^{0+\Delta x}-a^0}{\Delta x}=a^x\frac d{dx}(a^x)|_{x=0} \end{align*}$$ This means that to find the derivative, we must already know the derivative.

$\displaystyle e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$, so $\displaystyle\lim_{\Delta x\to0}\frac{e^{\Delta x}-1}{\Delta x} =\lim_{\Delta x\to0}\frac{\displaystyle\lim_{n\to\infty}\left(1+\frac {\Delta x}n\right)^n-1}{\Delta x}$.

Applying the binomial theorem, we get $\displaystyle\frac d{dx}(e^x)|_{x=0} =\lim_{\Delta x\to0}\frac{e^{\Delta x}-1}{\Delta x}\\=\lim_{\Delta x\to0}\frac{\displaystyle\lim_{n\to\infty}\left(1+\frac {\Delta x}n\right)^n-1}{\Delta x} \\=\lim_{\Delta x\to0}\frac{\displaystyle\lim_{n\to\infty}\left(1+\frac{\Delta x}{n} \cdot n + ... - 1\right)}{\Delta x}=1.$

Using this, we can find that $\displaystyle\frac d{dx}(e^x)=e^x\frac d{dx}(e^x)|_{x=0}=e^x$.

By manipulating the exponent, we now can get $a^x=(e^{\ln a})^x=e^{x\ln a}$.

Applying the Chain Rule, $$\begin{align*} \frac d{dx}(a^x)=\frac d{dx}(e^{x\ln a})&=e^{x\ln a}\frac d{dx}(x\ln a)\\\ &=a^x\ln a \end{align*}$$

Here is an interesting problem in elementary calculus:(taken from a problem solving group on Facebook.)

$f:[0,1]\rightarrow \textbf{R}$ is continuous on $[0,1]$ and differentiable on $(0,1)$ so that $f(0)=0$ and $0\leq f'(x)\leq k f(x)$, where $k\in \textbf{R}^{+}, \ \forall x \in (0,1)$.

Prove that $f(x)=0 \ \forall x \in (0,1)$.

(Video) Solving Every Derivative Ever

by PatrickJMT

Okay, maybe it's not EVERY derivative ever but here's a video by patrickJMT going over different derivatives for a whopping 35 minutes.

Let's have a season recap of some formulas before we go in. $$\begin{alignat}{2} & \frac{d}{dx}x^n=nx^{n-1} \quad && \frac{d}{dx}\cot x=-\csc^2 x\\ & \frac{d}{dx}k=0 \quad && \frac{d}{dx}\sec x=\sec x \tan x\\ & \frac{d}{dx}\sin x=\cos x \quad && \frac{d}{dx}e^x=e^x\\ & \frac{d}{dx}\cos x=-\sin x \quad && \frac{d}{dx}a^x=a^x\ln a\\ &\frac{d}{dx}\tan x=\sec^2x \quad && \frac{d}{dx}\ln x=\frac1x\\ &\frac{d}{dx}\csc x=-\csc x\cot x \quad &&\frac{d}{dx}\log_ax=\frac1{x\ln a} \end{alignat}$$

Also let's recap the product, quotient, and chain rule. $$\begin{align} [f(x)g(x)]'&=f'(x)g(x)+f(x)g'(x)\\ \bigg[\frac{f(x)}{g(x)}\bigg]'&=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\\ [f(\ g(x)\ )]'&=f'(\ g(x)\ )\cdot g'(x) \end{align}$$

Problem 1

$$\begin{align} y&=x^2+3x+4\\ y'&=2x^{2-1}+3(1x^{1-1})+0\\ &=\boxed{2x+3} \end{align}$$

Problem 2

$$\begin{align} y&=5t^{-\frac23}\\ y'&=5(-\frac23t^{-\frac23-1})\\ &=-\frac{10}3t^{-\frac53}\\ &=\boxed{-\frac{10}{3t^{\frac53}}} \end{align}$$

Problem 3

$$\begin{align} y&=\frac{3x-1}{\sqrt x}\\ &=\frac{3x-1}{x^{\frac12}}\\ &=\frac{3x}{x^{\frac12}}-\frac{1}{x^{\frac12}}\\ &=3x^{\frac12}-x^{-\frac12}\\ y'&=3(\frac12x^{-\frac12})-(-\frac12x^{-\frac32})\\ &=\frac{3}{2x^{\frac12}}+\frac{1}{2x^{\frac32}}\\ &=\boxed{\frac{3x+1}{2x^{\frac32}}} \end{align}$$

Problem 4

$$\begin{align} y&=(x+4)(x-4)\\ &=x^2-4x+4x-16\\ &=x^2-16\\ y'&=2x^{2-1}-0\\ &=\boxed{2x} \end{align}$$

Problem 5

$$\begin{align} y&=\frac{\sqrt x-1}{\sqrt x+1}\\ &=\frac{x^{\frac12}-1}{x^{\frac12}+1}\\ y'&=\frac{\color{red}{(x^{\frac12}-1)'}(x^{\frac12}+1)-(x^{\frac12}-1)\color{red}{(x^{\frac12}+1)'}}{(x^{\frac12}+1)^2}\\ &=\frac{\frac12x^{-\frac12}(x^{\frac12}+1)-(x^{\frac12}-1)(\frac12x^{-\frac12})}{(x^{\frac12}+1)^2}\\ &=\frac{\frac12x^{-\frac12}[(x^{\frac12}+1)-(x^{\frac12}-1)]}{(x^{\frac12}+1)^2}\\ &=\frac{(x^{\frac12}+1)-(x^{\frac12}-1)}{2\sqrt x(x^{\frac12}+1)^2}\\ &=\frac{2}{2\sqrt x(x^{\frac12}+1)^2}\\ &=\boxed{\frac{1}{\sqrt x(x^{\frac12}+1)^2}} \end{align}$$

Problem 6

$$\begin{align} y&=\frac{2x-\sqrt{x}\cdot x}{x^2}\\ &=\frac{2x-x^{\frac32}}{x^2}\\ &=\frac{2x}{x^2}-\frac{x^{\frac32}}{x^2}\\ &=\frac{2}{x}-\frac{1}{x^{\frac12}}\\ &=2x^{-1}-x^{-\frac12}\\ y'&=2(-1x^{-2})-(-\frac12x^{\frac32})\\ &=\boxed{\frac{1}{2x^{\frac32}}-\frac{2}{x^2}} \end{align}$$

Problem 7

$$\begin{align} y&=(\sin x)(x^{\frac12})\\ y'&=\color{red}{(\sin x)'}(x^{\frac12})+(\sin x)\color{red}{(x^{\frac12})'}\\ &=(\cos x)(x^{\frac12})+(\sin x)(\frac12x^{-\frac12})\\ &=\boxed{\sqrt{x}\cos x+\frac{\sin x}{2\sqrt x}} \end{align}$$

Problem 8

$$\begin{align} y&=(\sin x)(\cos x)\\ y'&=\color{red}{(\sin x)'}(\cos x)+(\sin x)\color{red}{(\cos x)}\\ &=(\cos x)(\cos x)+(\sin x)(-\sin x)\\ &=\cos^2x-\sin^2x\\ &=\boxed{\cos(2x)} \end{align}$$

Problem 9

$$\begin{align} y&=\sin(\cos x)\\ y'&=\cos(\cos x)\cdot\color{red}{[\cos x]'}\\ &=\cos(\cos x)\cdot(-\sin x)\\ &=\boxed{-\sin x\cos(\cos x)} \end{align}$$

Problem 10

$$\begin{align} y&=\sec[(1+x^2)^{\frac12}]\\ y'&=\sec[(1+x^2)^{\frac12}]\cdot\tan[(1+x^2)^{\frac12}]\cdot\color{red}{[(1+x^2)^{\frac12}]'}\\ &=\sec[(1+x^2)^{\frac12}]\cdot\tan[(1+x^2)^{\frac12}]\cdot\frac12(1+x^2)^{-\frac12}\cdot\color{red}{(1+x^2)'}\\ &=\sec[(1+x^2)^{\frac12}]\cdot\tan[(1+x^2)^{\frac12}]\cdot\frac12(1+x^2)^{-\frac12}\cdot(2x)\\ &=\boxed{\frac{x\sec[(1+x^2)^{\frac12}]\tan[(1+x^2)^{\frac12}]}{\sqrt{1+x^2}}} \end{align}$$

Problem 11

$$\begin{align} y&=\tan^2[\sin(\theta^3)]\\ y'&=2\tan[\sin(\theta^3)]\cdot\color{red}{[\ \tan[\sin(\theta^3)]\ ]'}\\ &=2\tan[\sin(\theta^3)]\cdot\sec^2[\sin(\theta^3)]\cdot\color{red}{[\sin(\theta^3)]'}\\ &=2\tan[\sin(\theta^3)]\cdot\sec^2[\sin(\theta^3)]\cdot\cos(\theta^3)\cdot\color{red}{(\theta^3)'}\\ &=2\tan[\sin(\theta^3)]\cdot\sec^2[\sin(\theta^3)]\cdot\cos(\theta^3)\cdot3\theta^2\\ &=\boxed{6\theta^2\tan[\sin(\theta^3)]\sec^2[\sin(\theta^3)]\cos(\theta^3)} \end{align}$$

Problem 12

$$\begin{align} y&=x\sin(3x^{-1})\\ y'&=\color{red}{(x)'}[\sin(3x^{-1})]+(x)\color{red}{[\sin(3x^{-1})]'}\\ &=(1)\sin(3x^{-1})+(x)\cos(3x^{-1})\cdot\color{red}{(3x^{-1})'}\\ &=\sin(3x^{-1})+(x)\cos(3x^{-1})\cdot3(-1x^{-2})\\\\ &=\sin(3x^{-1})-\frac{3x\cos(3x^{-1})}{x^{2}}\\ &=\sin(\frac3x)-\frac{3\cos(\frac3x)}{x}\\ &=\boxed{\frac{x\sin(\frac3x)-3\cos(\frac3x)}{x}} \end{align}$$

Problem 13

$$\begin{align} y&=\frac{(x^3+4)^2}{(x^2+1)^4}\\ y'&=\frac{\color{red}{[(x^3+4)^2]'}(x^2+1)^4-(x^3+4)^2\color{red}{[(x^2+1)^4]'}}{[(x^2+1)^4]^2}\\ &=\frac{2(x^3+4)\color{red}{(x^3+4)'}(x^2+1)^4-4(x^3+4)^2(x^2+1)^3\color{red}{(x^2+1)'}}{(x^2+1)^8}\\ &=\frac{6x^2(x^3+4)(x^2+1)^4-8x(x^3+4)^2(x^2+1)^3}{(x^2+1)^8}\\ &=\frac{6x^2(x^3+4)(x^2+1)^4-8x(x^3+4)^2(x^2+1)^3}{(x^2+1)^8}\\ &=\frac{2x(x^3+4)(x^2+1)^3[3x(x^2+1)-4(x^3+4)]}{(x^2+1)^8}\\ &=\frac{2x(x^3+4)[3x(x^2+1)-4(x^3+4)]}{(x^2+1)^5}\\ &=\frac{2x(x^3+4)[(3x^3+3x)-(4x^3+16)]}{(x^2+1)^5}\\ &=\boxed{\frac{2x(x^3+4)(-x^3+3x-16)}{(x^2+1)^5}} \end{align}$$

Problem 14

$$\begin{align} y&=\ln(x)+x^3\\ y'&=\frac1x+3x^2\\ &=\boxed{\frac{3x^3+1}{x}} \end{align}$$

Problem 15

$$\begin{align} y&=x^2e^x\\ y'&=\color{red}{(x^2)'}e^x+x^2\color{red}{(e^x)'}\\ &=2xe^x+x^2e^x\\ &=\boxed{xe^x(x+2)} \end{align}$$

Problem 16

$$\begin{align} y&=e^{t\cdot\cos(2t)}\\ y'&=e^{t\cdot\cos(2t)}\cdot\color{red}{[t\cdot\cos(2t)]'}\\ &=e^{t\cdot\cos(2t)}\cdot[\color{red}{(t)'}\cos(2t)+t\color{red}{[\cos(2t)]'}]\\ &=e^{t\cdot\cos(2t)}\cdot[\cos(2t)-t\sin(2t)\cdot\color{red}{(2t)'}]\\ &=\boxed{e^{t\cdot\cos(2t)}\cdot[\cos(2t)-2t\sin(2t)]} \end{align}$$

Problem 17

$$\begin{align} y&=\cos(e^{\pi x})\\ y'&=-\sin(e^{\pi x})\cdot\color{red}{(e^{\pi x})'}\\ &=-\sin(e^{\pi x})\cdot e^{\pi x}\color{red}{(\pi x)'}\\ &=\boxed{-\pi e^{\pi x}\sin(e^{\pi x})} \end{align}$$

Problem 18

$$\begin{align} y&=(1+2e^{3x})^{\frac12}\\ y'&=\frac12(1+2e^{3x})^{-\frac12}\cdot\color{red}{(1+2e^{3x})'}\\ &=\frac12(1+2e^{3x})^{-\frac12}\cdot2e^{3x}\cdot\color{red}{(3x)'}\\ &=\boxed{\frac{3e^{3x}}{\sqrt{1+2e^{3x}}}} \end{align}$$

Problem 19

$$\begin{align} y&=5^{-\frac1x}\\ &=5^{-x^{-1}}\\ y'&=5^{-x^{-1}}\cdot\ln5\cdot\color{red}{(-x^{-1})'}\\ &=5^{-x^{-1}}\cdot\ln5\cdot(x^{-2})\\ &=\boxed{\frac{\ln5}{5^{\frac1x}x^2}} \end{align}$$

Problem 20

$$\begin{align} y&=\ln(x^4\sin^2x)\\ y'&=\frac{1}{x^4\sin^2x}\cdot\color{red}{[x^4\sin^2x]'}\\ &=\frac{1}{x^4\sin^2x}\cdot[\color{red}{(x^4)'}\sin^2x+x^4\color{red}{(\sin^2x)'}]\\ &=\frac{4x^3\sin^2x+2x^4\sin x\cdot\color{red}{(\sin x)'}}{x^4\sin^2x}\\ &=\frac{4x^3\sin^2x+2x^4\sin x\cos x}{x^4\sin^2x}\\ &=\frac{2x^3\sin x(2\sin x+x\cos x)}{x^4\sin^2x}\\ &=\frac{2(2+x\cos x)}{x\sin x}\\ &=\boxed{\frac4x+2\cot x} \end{align}$$

Problem 21

$$\begin{align} y&=\ln\sqrt{\frac{5x-1}{2x+3}}\\ &=\ln\bigg(\frac{5x-1}{2x+3}\bigg)^{\frac12}\\ &=\frac12\bigg[\ln\bigg(\frac{5x-1}{2x+3}\bigg)\bigg]\\ &=\frac12[\ln(5x-1)-\ln(2x+3)]\\ y'&=\frac12[\frac{1}{5x-1}\cdot\color{red}{(5x-1)'}-\frac{1}{2x+3}\cdot\color{red}{(2x+3)'}]\\ &=\frac12\bigg[\frac{5}{5x-1}-\frac{2}{2x+3}\bigg]\\ &=\frac12\bigg[\frac{5(2x+3)-2(5x-1)}{(5x-1)(2x+3)}\bigg]\\ &=\frac12\bigg[\frac{10x+15--10x+2}{10x^2+13x-3}\bigg]\\ &=\frac12\bigg[\frac{17}{10x^2+13x-3}\bigg]\\ &=\boxed{\frac{17}{20x^2+26x-6}} \end{align}$$

Problem 22

$$\begin{align} y&=t^3-3^t\\ y'&=3t^2-3^t\cdot\ln3\\ &=\boxed{3t^2-3^t\ln3} \end{align}$$

Problem 23

$$\begin{align} y&=\ln(\sec x+\tan x)\\ y'&=\frac{1}{\sec x+\tan x}\cdot\color{red}{[\sec x+\tan x]'}\\ &=\frac{1}{\sec x+\tan x}\cdot[\sec x\cdot\tan x+\sec^2x]\\ &=\boxed{\frac{\sec x\tan x+\sec^2x}{\sec x+\tan x}} \end{align}$$

Problem 24

$$\begin{align} y&=\log_{10}\bigg(\frac{2x}{x-1}\bigg)\\ &=\log_{10}(2x)-\log_{10}(x-1)\\ y'&=\frac{1}{2x\ln10}\cdot\color{red}{(2x)'}-\frac{1}{(x-1)\ln10}\cdot\color{red}{(x-1)'}\\ &=\frac{2}{2x\ln10}-\frac{1}{(x-1)\ln10}\\ &=\boxed{\frac{1}{x\ln10}-\frac{1}{(x-1)\ln10}} \end{align}$$

Problem 25

$$\begin{align} y&=\frac{\ln x}{x^2}\\ y'&=\frac{\color{red}{(\ln x)'}(x^2)-(\ln x)\color{red}{(x^2)'}}{[x^2]^2}\\ &=\frac{\frac1x(x^2)-2x(\ln x)}{x^4}\\ &=\frac{x-2x(\ln x)}{x^4}\\\\ &=\boxed{\frac{1-2\ln x}{x^3}} \end{align}$$

(Video) Derivatives of Exponential Functions

by PatrickJMT

Here's a video by patrickJMT showing you how to find the derivative of exponential functions. The two main rules are: $$\frac{d}{dx}e^{f(x)}=e^{f(x)}\cdot f'(x)\\ \frac{d}{dx}a^{f(x)}=a^{f(x)}\cdot f'(x)\cdot \ln a$$

The first example he finds the derivative of is: $$f(x)=4^x+3e^x+x^4$$ This is a simple problem where you just have to use the rules to solve. $$\boxed{f'(x)=4^x\cdot\ln 4+3e^x+4x^3}$$

The second example is: $$g(x)=2e^{\cos x\cdot\sin(5x)}$$ This one is a little bit trickier. We have a product rule inside a chain rule. Other than that, the derivative of the exponent part is super easy. $$\begin{align} g'(x)&=2e^{\cos x[\sin(5x)]}\cdot\color{red}{[\cos x\cdot\sin(5x)]'}\\ &=2e^{\cos x[\sin(5x)]}\cdot[\color{red}{[\cos x]'}\sin(5x)+\cos x\color{red}{[\sin(5x)]'}]\\ &=2e^{\cos x[\sin(5x)]}\cdot[-\sin x\cdot\sin(5x)+\cos x\cdot\cos(5x)\cdot\color{red}{(5x)'}]\\ &=\boxed{2e^{\cos x[\sin(5x)]}\cdot[5\cos x\cdot\cos(5x)-\sin x\cdot\sin(5x)]}\\ \end{align}$$

(Videos) Derivatives of Exponential Functions

by PatrickJMT

Here's a video by PatrickJMT working through problems on finding the derivative of exponential functions.

The two derivative rules are for $e$ raised to a function or a number raised to a function. $$\frac{d}{dx}e^{f(x)}=e^{f(x)}\cdot f'(x)\\ \frac{d}{dx}a^{f(x)}=a^{f(x)}\cdot\ln a\cdot f'(x)$$

The first example he solves is finding the derivative of: $$f(x)=e^xx^2$$ Here we have a product rule. $$\begin{align} f'(x)&=\color{red}{(e^x)'}x^2+e^x\color{red}{(x^2)'}\\ &=\boxed{e^xx^2+2xe^x} \end{align}$$

The second example is: $$f(x)=e^{x\sin(2x)}$$ In this problem there's a product rule inside a chain rule. $$\begin{align} f'(x)&=e^{x\sin(2x)}\cdot\color{red}{[x\sin(2x)]'}\\ &=e^{x\sin(2x)}\cdot[\color{red}{(x)'}\sin(2x)+x\color{red}{[\sin(2x)]'}]\\ &=e^{x\sin(2x)}\cdot[\sin(2x)+x\cdot\cos(2x)\cdot\color{red}{(2x)'}\\ &=\boxed{e^{x\sin(2x)}[\sin(2x)+2x\cos(2x)]} \end{align}$$

The final example he tackles is: $$f(t)=\cos(2^{\pi t})$$ Here the exponential function is inside a chain rule. $$\begin{align} f'(t)&=-\sin(2^{\pi t})\cdot\color{red}{[2^{\pi t}]'}\\ &=-\sin(2^{\pi t})\cdot2^{\pi t}\cdot\ln2\cdot\color{red}{(\pi t)'}\\ &=-\sin(2^{\pi t})\cdot2^{\pi t}\cdot\ln2\cdot\pi\\ &=\boxed{-\pi\ln(2)2^{\pi t}\sin(2^{\pi t})} \end{align}$$

Here's one more video where he goes over a different kind of derivative problem. It asks:

For what values of $x$ does the function $h(x)=5e^{5x}-25x$ have negative derivatives?

So what this problem is asking is to solve for when the derivative is less than zero. $$h'(x)<0$$

We start by finding the derivative of our function. $$\begin{align} h'(x)&=5e^{5x}\cdot\color{red}{(5x)'}-25\\ &=5e^{5x}\cdot5-25\\ &=25(e^{5x}-1) \end{align}$$

Now we need to solve our inequality, $h'(x)<0$. $$\begin{align} 25(e^{5x}-1)&<0\\ e^{5x}-1&<0\\ e^{5x}&<1\\ \ln(e^{5x})&<\ln 1\\ 5x\ln e&<0\\ 5x&<0\\ x&<0 \end{align}$$

So when $x$ is between $(-\infty,0)$ then the derivative, $h'(x)$, is negative.