Instructions
You are provided with a function of the form f(x) = axⁿ, that consists of a single term only and 'a' and 'n' are integers, e.g f(x) = 3x², f(x) = 5 etc.
Your task is to create a function that takes f(x) as the argument and returns the result of differentiating the function, that is, the derivative.
If f(x)=axn f(x) = ax^nf(x)=ax
n
, then f′(x)=naxn−1 f^{prime}(x) = nax^{n-1}f
′
(x)=nax
n−1
Input is a string, for example “5x^4”. The function f(x) consists of a single term only. Variable is denoted by x.
Output should be a string, for example “20x^3”.
Examples
“3x^2” => “6x”
“-5x^3” => “-15x^2”
“6x^-2” => “-12x^-3”
“5x” => “5”
“-x” => “-1”
“42” => “0”
My solution:
function differentiate(f) { if(!f.includes('x')) return "0" if(f == 'x') return "1" if(f == '-x') return "-1" if(!f.includes('^')) return f.replace('x', '') f=f.split('^') if(f[0] == 'x' || f[0] == '-x') f[0]=f[0].replace('x','1') f[0] = f[0].replace('x','') let exponent = +f[1] let base = +f[0] let newExponent = +f[1] - 1 let newBase = exponent*base if(newExponent == 1) return newBase + 'x' if(newBase == -1) return '-x^' + newExponent if(newBase == 1) return 'x^' + newExponent return newBase + 'x^' + newExponent } Explanation
First I used some conditionals for especific inputs, if the function doesn't includes an “x” it is just a number like “42”, so the result should be 0
if(!f.includes('x')) return “0”
If it is just “x” it'll return 1, and if it is “-x” it'll return -1
if(f == '-x') return “-1”
And if it doesn't includes a “^” it'll return the function but without the x, so if I have “6x” it'll return “6”
if(!f.includes('^')) return f.replace('x', '')
After that I splitted in the “^” and I used a conditional to see that if in the first element that is the base before the “^” I have only a “x” or a “-x” those are equal to 1 or -1, so I replaced “x” for 1 and “-x” for -1
f=f.split('^')
if(f[0] == 'x' || f[0] == '-x') f[0]=f[0].replace('x','1')
For example: x^2 –> [1, 2]
If not, I'll just eliminate the “x” replacing it to an empty string
f[0] = f[0].replace('x','')
After that using the array that I splitted, I declarated the exponent and the base, after that I created the newExponent resting 1 to the exponent and the newBase multiplying the base for the exponent.
let exponent = +f[1]
let base = +f[0]
let newExponent = +f[1] – 1
let newBase = exponent*base
Then I used some conditionals to return the last results, if the newExponent is 1, it should return the newBase and x, because if it is elevated to 1 it remains the same, if the newBase is equal to -1, it'll return “-x” and the newExponent, and if it is 1 it'll return “x” and the newExponent, and if any of this contions is true it'll return the newBase + 'x^' + newExponent.
if(newExponent == 1) return newBase + 'x'
if(newBase == -1) return '-x^' + newExponent
if(newBase == 1) return 'x^' + newExponent
return newBase + 'x^' + newExponent
What do you think about this solution? 👇🤔