No, the -1 on the inverse function is on the operation itself, while the -1 on the reciprocal functions is on the outside. eg: sin^-1(x) = arcsinx, (sin(x))^-1=cscx
There is no analytic solution. You can use Newton's method to narrow in on the solution. Shuffle everything to the left: 2^x + 3^x + 4^x - 5^x = 0 Let f(x) = 2^x + 3^x + 4^x - 5^x Take derivative: f'(x) = ln(2)*2^x + ln(3)*3^x + ln(4)*4^x - ln(5)*5^x Set up Newton's method, guessing x_0 = 2, and using it and its derivative as feedback to find the next x-value. x_nplus1 = x_n - f(x_n)/f'(x_n) n ____ x_n ____ f(x_n)_________f'(x_n) 0_____ 2 ________ 4 __________ -5.395 1_____ 2.741___-10.76 _______ -43.73 2_____ 2.496___-2.568 _______ -24.34 3_____ 2.391___-0.318 _______ -18.52 4_____ 2.374___-0.007 _______ -17.69 5_____ 2.373___-3.954E-06 ___ -17.66 Solution: about x = 2.373
Derivatives of ALL trig functions (proofs!)
th-cam.com/video/IuClHA9VjDY/w-d-xo.html
2 (co)secants and 1 (co)tangent
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thanks for the math!
thanks sir
what about the inverse of these functions?
Isn't the inverse alr there?
Like sin^-1 = csc
cos^-1 = sec
tan^-1 = cot
No, the -1 on the inverse function is on the operation itself, while the -1 on the reciprocal functions is on the outside. eg: sin^-1(x) = arcsinx, (sin(x))^-1=cscx
Can you please solve 2^x + 3^x + 4^x = 5^x
There is no analytic solution. You can use Newton's method to narrow in on the solution.
Shuffle everything to the left:
2^x + 3^x + 4^x - 5^x = 0
Let f(x) = 2^x + 3^x + 4^x - 5^x
Take derivative:
f'(x) = ln(2)*2^x + ln(3)*3^x + ln(4)*4^x - ln(5)*5^x
Set up Newton's method, guessing x_0 = 2, and using it and its derivative as feedback to find the next x-value.
x_nplus1 = x_n - f(x_n)/f'(x_n)
n ____ x_n ____ f(x_n)_________f'(x_n)
0_____ 2 ________ 4 __________ -5.395
1_____ 2.741___-10.76 _______ -43.73
2_____ 2.496___-2.568 _______ -24.34
3_____ 2.391___-0.318 _______ -18.52
4_____ 2.374___-0.007 _______ -17.69
5_____ 2.373___-3.954E-06 ___ -17.66
Solution: about x = 2.373