Swift 6, 218 214 175 bytes

{""+($0+"").split(separator:" + ").flatMap{let p=$0.split{$0>"]"},d=p.count>1 ?Int(p[1])!:1
return p[0]<$0 ?"\(d*Int(p[0])!)\(d<2 ?"":d<3 ?"x":"x^\(d-1)") + ":""}.dropLast(3)}

Try it on SwiftFiddle!

Perl, 64 63 bytes

62b code + 1 command line (-p)

s/(\d+)x.(\d+)/$1*$2."x^".($2-1)/eg;s/\^1\b|^\d+ . |x(?!\^)//g

Usage example:

echo "1 + 2x + 3x^2" | perl -pe 's/(\d+)x.(\d+)/$1*$2."x^".($2-1)/eg;s/\^1\b|^\d+ . |x(?!\^)//g'

Thanks Denis for -1b

Python 2, 174 bytes

print' + '.join(['%d%s%s'%(b[0]*b[1],'x'*(b[1]>1),'^%d'%(b[1]-1)*(b[1]>2))for b in[map(int,a.split('x^')if 'x^'in a else[a[:-1],1])for a in input().split(' + ')if 'x'in a]])

Unfortunately, DLosc's trick to rename the split method and perform the differentiation in a specific function does not shorten my code...

Python 2, 229 bytes

import os
print' + '.join([i,i[:-2]][i[-2:]=='^1'].replace('x^0','')for i in[`a*b`+'x^'+`b-1`for a,b in[map(int,a.split('x^'))for a in[[[i+'x^0',i+'^1'][i[-1]=='x'],i]['^'in i]for i in os.read(0,9**9).split(' + ')]]]if i[0]!='0')

CJam, 43 41 bytes

Qleu'^/';*'+/{~:E[*'x['^E(]]E<}/]1>" + "*

Thanks to @jimmy23013 for pointing out one bug and golfing off two bytes!

Try it online in the CJam interpreter.

How it works

Q           e# Leave an empty array on the bottom of the stack.
l           e# Read a line from STDIN.
eu'^/';*    e# Convert to uppercase and replace carets with semicolons.
'+/         e# Split at plus signs.

{           e# For each resulting chunk:
  ~         e#   Evaluate it. "X" pushes 1 and ";" discards.
            e#   For example, "3X" pushes (3 1) and "3X;2" (3 2).
   :E       e#   Save the rightmost integer (usually the exponent) in E.
   [        e#
     *      e#   Multiply both integers.
            e#   For a constant term c, this repeats the empty string (Q) c times.
     'x     e#   Push the character 'x'.
     ['^E(] e#   Push ['^' E-1].
   ]        e#
   E<       e#  Keep at most E elements of this array.
            e#  If E == 1, 'x' and ['^' E-1] are discarded.
            e#  If E == 2, ['^' E-1] is discarded.
            e#  If E >= 3, nothing is discarded.
}/          e#

]           e# Wrap the entire stack in an array.
1>          e# Discard its first element.
            e# If the first term was constant, this discards [""]. ["" 'x']
            e# or ["" 'x' ['^' E-1]], depending on the constant.
            e# In all other cases, it discards the untouched empty array (Q).
" + "*      e# Join all kept array elements, separating by " + ".

Python 3, 176 bytes

s=input().split(' + ')
y='x'in s[0]
L=map(lambda x:map(int,x.split('x^')),s[2-y:])
print(' + '.join([s[1-y][:-1]]+['x^'.join(map(str,[a*b,b-1])).rstrip('^1')for a,b in L]))

Indeed, the main annoyance is having to convert between strings and ints. Also, if a constant term was required, the code would only be 153 bytes.

CJam, 62 57 55 49 bytes

Well, Dennis put this to shame before I even noticed that the site was back up. But here is my creation anyway:

lS/{'x/:T,({T~1>:T{~T~*'xT~(:T({'^T}&}&" + "}&}/;

Latest version saves a few bytes with shortcuts suggested by @Dennis (use variables, and split at space instead of +).

Try it online

Python 2, 166 bytes

Boy, this seems longer than it should be.

S=str.split
def d(t):e="^"in t and int(S(t,"^")[1])-1;return`int(S(t,"x")[0])*(e+1)`+"x"[:e]+"^%d"%e*(e>1)
print" + ".join(d(t)for t in S(raw_input()," + ")if"x"in t)

The function d takes a non-constant term t and returns its derivative. The reason I def the function instead of using a lambda is so I can assign the exponent minus 1 to e, which then gets used another four times. The main annoying thing is having to cast back and forth between strings and ints, although Python 2's backtick operator helps with that.

We then split the input into terms and call d on each one that has "x" in it, thereby eliminating the constant term. The results are joined back together and printed.

C, 204 162 bytes

#define g getchar()^10
h,e;main(c){for(;!h&&scanf("%dx%n^%d",&c,&h,&e);h=g?g?e?printf(" + "):0,0:1:1)e=e?e:h?1:0,e?printf(e>2?"%dx^%d":e>1?"%dx":"%d",c*e,e-1):0;}

Basically parse each term and print out the differentiated term in sequence. Fairly straightforward.

Retina, 53 43 42 41 40 35 bytes

^[^x]+ |(\^1)?\w(?=1*x.(1+)| |$)
$2

For counting purposes each line goes in a separate file, but you can run the above as a single file by invoking Retina with the -s flag.

This expects the numbers in the input string to be given in unary and will yield output in the same format. E.g.

1 + 11x + -111x^11 + 11x^111 + -1x^11111
-->
11 + -111111x + 111111x^11 + -11111x^1111

instead of

1 + 2x + -3x^2 + 2x^3 + -1x^5
-->
2 + -6x + 6x^2 + -5x^4

Explanation

The code describes a single regex substitution, which is basically 4 substitutions compressed into one. Note that only one of the branches will fill group $2 so if any of the other three match, the match will simply be deleted from the string. So we can look at the four different cases separately:

^[^x]+<space>
<empty>

If it's possible to reach a space from the beginning of the string without encountering an x that means the first term is the constant term and we delete it. Due to the greediness of +, this will also match the plus and the second space after the constant term. If there is no constant term, this part will simply never match.

x(?= )
<empty>

This matches an x which is followed by a space, i.e. the x of the linear term (if it exists), and removes it. We can be sure that there's a space after it, because the degree of the polynomial is always at least 2.

1(?=1*x.(1+))
$1

This performs the multiplication of the coefficient by the exponent. This matches a single 1 in the coefficient and replaces it by the entire corresponding exponent via the lookahead.

(\^1)?1(?= |$)
<empty>

This reduces all remaining exponents by matching the trailing 1 (ensured by the lookahead). If it's possible to match ^11 (and a word boundary) we remove that instead, which takes care of displaying the linear term correctly.

For the compression, we notice that most of the conditions don't affect each other. (\^1)? won't match if the lookahead in the third case is true, so we can put those two together as

(\^1)?1(?=1*x.(1+)| |$)
$2

Now we already have the lookahead needed for the second case and the others can never be true when matching x, so we can simply generalise the 1 to a \w:

(\^1)?\w(?=1*x.(1+)| |$)
$2

The first case doesn't really have anything in common with the others, so we keep it separate.

Julia, 220 bytes

No regular expressions!

y->(A=Any[];for i=parse(y).args[2:end] T=typeof(i);T<:Int&&continue;T==Symbol?push!(A,1):(a=i.args;c=a[2];typeof(a[3])!=Expr?push!(A,c):(x=a[3].args[3];push!(A,string(c*x,"x",x>2?string("^",ex-1):""))))end;join(A," + "))

This creates a lambda function that accepts a string and returns a string. The innards mimic what happens when Julia code is evaluated: a string is parsed into symbols, expressions, and calls. I might actually try writing a full Julia symbolic differentiation function and submit it to be part of Julia.

Ungolfed + explanation:

function polyderiv{T<:AbstractString}(y::T)

    # Start by parsing the string into an expression
    p = parse(y)

    # Define an output vector to hold each differentiated term
    A = Any[]

    # Loop through the elements of p, skipping the operand
    for i in p.args[2:end]

        T = typeof(i)

        # Each element will be an integer, symbol, or expression.
        # Integers are constants and thus integrate to 0. Symbols
        # represent x alone, i.e. "x" with no coefficient or
        # exponent, so they integrate to 1. The difficulty here
        # stems from parsing out the expressions.

        # Omit zero derivatives
        T <: Int && continue

        if T == Symbol
            # This term will integrate to 1
            push!(A, 1)
        else
            # Get the vector of parsed out expression components.
            # The first will be a symbol indicating the operand,
            # e.g. :+, :*, or :^. The second element is the
            # coefficient.
            a = i.args

            # Coefficient
            c = a[2]

            # If the third element is an expression, we have an
            # exponent, otherwise we simply have cx, where c is
            # the coefficient.
            if typeof(a[3]) != Expr
                push!(A, c)
            else
                # Exponent
                x = a[3].args[3]

                # String representation of the differentiated term
                s = string(c*x, "x", x > 2 ? string("^", x-1) : "")

                push!(A, s)
            end
        end
    end

    # Return the elements of A joined into a string
    join(A, " + ")
end

JavaScript ES6, 108 bytes

f=s=>s.replace(/([-\d]+)(x)?\^?(\d+)?( \+ )?/g,(m,c,x,e,p)=>x?(c*e||c)+(--e?x+(e>1?'^'+e:''):'')+(p||''):'')

ES5 Snippet:

// ES5 version, the only difference is no use of arrow functions.
function f(s) {
  return s.replace(/([-\d]+)(x)?\^?(\d+)?( \+ )?/g, function(m,c,x,e,p) {
    return x ? (c*e||c) + (--e?x+(e>1?'^'+e:''):'') + (p||'') : '';
  });
}

[
  '3 + 1x + 2x^2',
  '1 + 2x + -3x^2 + 17x^17 + -1x^107',
  '17x + 1x^2'
].forEach(function(preset) {
  var presetOption = new Option(preset, preset);
  presetSelect.appendChild(presetOption);
});

function loadPreset() {
  var value = presetSelect.value;
  polynomialInput.value = value;
  polynomialInput.disabled = !!value;
  showDifferential();
}

function showDifferential() {
  var value = polynomialInput.value;
  output.innerHTML = value ? f(value) : '';
}

code {
  display: block;
  margin: 1em 0;
}

<label for="presetSelect">Preset:</label>
<select id="presetSelect" onChange="loadPreset()">
  <option value="">None</option>
</select>
<input type="text" id="polynomialInput"/>
<button id="go" onclick="showDifferential()">Differentiate!</button>
<hr />
<code id="output">
</code>

Pyth, 62 bytes

jJ" + "m::j"x^",*Fdted"x.1$"\x"x.0"kftTmvMcd"x^"c:z"x ""x^1 "J

Pretty ugly solution, using some regex substitutions.