General Differential Calculus Formulas, and Functions

Inverse Trigonometric Functions
arcsine x = sin-1 x = y --> sin y = x
arccos x = cos-1 x = y --> cos y = x
arctan x = tan-1 x = y --> tan y = x

Exponential and Logarithimic Functions
log(a) = y --> a^y = x
ln(x) = log(e)x where x is greater than 0
ln(x) = y --> e^y = x

Cancellation Equations
loga(a^x) = x ln(e^x) = x
a^loga(x) = x e^ln(x) = x

Laws of logarithms
loga(xy) = loga(x) + loga(y)
loga(x/y) = loga(x) - loga(y)
loga(x^r) = rloga(x)

Hyperbolic Functions
sinh(x) = (e^x-e^-x)/2
cosh(x) = (e^x+e^-x)/2
tanh(x) = sinh(x)/cosh(x) = [(e^x-e^-x)/2] / [(e^x+e^-x)/2] = (e^x-e^-x) /(e^x+e^-x)
csch(x) = 1/sinh(x) = 2/(e^x-e^-x)
sech(x) = 1/cosh(x) = 2/(e^x+e^-x)
coth (x) = cosh (x)/sinh(x) = (e^x+e^-x) / (e^x-e^-x)

Inverse Hyperbolic Functions
y = sinh-1(x) --> sinh y = x
y = cosh-1(x) -->cosh y = x
y = tanh-1(x) -->tanh y = x
sinh-1 (x) = ln(x+squrt(x^2+1))
cosh-1 (x) = ln(x+squrt(x^2-1))
tanh-1 (x) = 1/2ln((1+x)/(1-x))

Differentiation Rules (where c is a constant number)
d/dx (c) = 0
Example: d/dx (2) = 0 and d/dx (2x) = 2

d/dx [f(x) + g(x)] = f'(x) + g'(x)
Example: d/dx (x^2 +x^3) = d/dx x^2 + d/dx x^3 = 2x+3x^2

d/dx [f(x) - g(x)] = f'(x) - g'(x)
Example: d/dx (x^2 - x^3) = d/dx x^2 - d/dx x^3 = 2x-3x^2

d/dx [f(x)*g(x)] = f(x)*g'(x) + g(x)*f'(x)
Example: d/dx [(x^2)*(x^3)] = [(x^2)*(3x^2) + (2x)*(x^3)] = 3x^4 + 2x^4 = 5x^4

d/dx [f(x)/g(x)] = [(g(x)*f'(x) - g'(x)*f(x))/(g(x))^2]
Example: d/dx [(x^2)/(x^3)] = [(x^3)*(2x) - (x^2)*(3x^2)/(x^3)^2] = 2x^4 - 3x^4/x^6 =
-x^4/x^6 = -1/x^2

d/dx f(g(x)) = f'(g(x))*g'(x)
Example: d/dx [squrt(x^2 + x^3)] = [1/(spurt(x^2 + x^3)) *(2x+3x^2)] =
[(2x+3x^2)/(squrt(x^2=x^3)]

d/dx [cf(x)] = c*f'(x)
Example: d/dx 2*(x^2) = 2*(2x) = 4x

d/dx (x^n) = n*x^(n-1)
Example: d/dx (x^3) = 3*(x^(3-1)) = 3x^2

Exponential and Logarithmic Functions
d/dx (e^x) = e^x
d/dx (a^x) = a^x*ln(a)
d/dx lnx = 1/x
d/dx (loga(x)) = [1/(a*lnx)]

Trigonometric Functions
d/dx (sin(x)) = cos(x)
d/dx (cos(x)) = -sin(x)
d/dx (tan(x)) = sec^2(x)
d/dx (csc(x)) = -csc(x)cot(x)
d/dx (sec(x)) = sec(x)tan(x)
d/dx (cot(x)) = -csc^2(x)

Inverse Trigonometric Functions
d/dx (sin-1(x)) = 1/(squrt(1-x^2))
Proof: Find d/dx (sin-1(x))
Sin-1(x) = y means sin(y) = x
d/dx sin(y) = d/dx (x)
cos(y)dy/dx = 1
dy/dx = 1/cos(y)
With the Pythagorean Theorem which says cos^2(x) + sin^2(x) = 1 or x^2 + y^2 = r^2
Therefore we can say: cos(x) = squrt(1-sin^2) = squrt(1-x^2)
So we have dy/dx = 1/[squrt(1-x^2)]

d/dx (cos-1(x)) = -1/(squrt(1-x^2))
Proof: find d/dx (cos-1(x))
cos-1(x) = y means cos(y) = x
d/dx cos(y) = d/dx (x)
-sin(y)dy/dx = 1
dy/dx = 1/-sin(y) = -1/sin(y)
With the Pythagorean Theorem which says cos^2(x) + sin^2(x) = 1 or x^2 + y^2 = r^2
Therefore we can say that sin(y) = squrt(1-cos^2(y)) = squrt(1-x^2)
So we have dy/dx = -1/[squrt(1-x^2)]

d/dx (tan-1(x)) = 1/(1+x^2)
Proof: find d/dx (tan-1(x))
(tan-1(x)) = y means tan(y) = x
d/dx (tan(y)) = d/dx (x)
sec^2(y)dy/dx = 1
dy/dx = 1/sec^2(y)
With the Pythagorean Theorem which says that 1 + tan^2(x) = sec^2(x) or 1 + x^2 = sec^2(x)
so we have dy/dx = 1/(1+x^2)

d/dx (csc-1(x)) = -1/[(x)(squrt(x^2-1))]
Proof: find d/dx (csc-1(x))
csc-1(x) = y means csc(y) = x
d/dx csc(y) = d/dx (x)
-cot(y)csc(y)dy/dx = 1
dy/dx = 1/-(cot(y)csc(y)) = -1/cot(y)csc(y)
Since 1+ cot^2(y) = csc^2(y)
Then cot^2(y) = csc^2(y) - 1
cot(y) = (squrt(csc^2(y)-1) = (squrt(x^2-1))
And csc^2(y) = 1 + cot^2(y)
So csc(y) = squrt(1 + cot^2(y)) = squrt(1 + x^2)
Therfore dy/dx = -1/cot(y)csc(y) = -1/[(squrt(x^2-1))*(squrt(1 + x^2))]
= -1/[x(squrt(x^2-1))]

d/dx (sec-1(x)) = 1/[(x)(squrt(x^2-1))]
Proof: find d/dx (sec-1(x))
sec-1(x) = y means sec(y) = x
d/dx sec(y) = d/dx (x)
tan(y)sec(y)dy/dx = 1
dy/dx = 1/(tan(y)sec(y))
Since 1+tan^2(y) = sec^2(y) then sec(y) = (squrt(1+ tan^(y)) = squrt(1+x^2)
and tan(y) = squrt(sec^2(y)-1) = squrt(x^2-1)
the 1/(tan(y)sec(y)) = 1/[squrt(1+x^2)squrt(x^2-1)] = 1/[(x)(squrt(x^2-1))]

d/dx (cot-1(x)) = -1/(1+x^2)
Proof: find d/dx (cot-1(x))
cot-1(x) = y means cot(y) = x
d/dx cot(y) = d/dx (x)
-csc^2(y)dy/dx = 1
dy/dx = 1/-csc^2(y) = -1/csc^2(y)
Since 1 + cot^2(y) = csc^2(y) = (squrt(1+x^2))
Then dy/dx = -1/csc^2(y) = -1/(squrt(1+x^2))

Hyperbolic Functions
d/dx (sinh(x)) = cosh(x)
d/dx (cosh(x)) = sinh(x)
d/dx (tanh(x)) = sec^2(x)
d/dx (csch(x)) = -csch(x)coth(x)
d/dx (sech(x)) = -sech(x)tanh(x)
d/dx (coth(x)) = -csch^2(x)

Inverse Hyperbolic Functions
d/dx (sinh-1(x)) = 1/(squrt(1+x^2))
d/dx (cosh-1(x)) = 1/(squrt(x^2-1))
d/dx (tanh-1(x)) = 1/(1-x^2)
d/dx (csch-1(x)) = -1/[x(squrt(x^2+1)]
d/dx (sech-1(x)) = -1/(x)[(spurt(1-x^2))]
d/dx (coth-1(x)) = 1/(1-x^2)

Implicit differentiation
If we have an equation x² + y² = 25 and we want to find dy/dx (the derivative of y in terms of x) we can use implicit differentiation.

Example
Given: x² + y² = 25 and we want to find the equation to the tangent line at the point (3,4) and we know dy/dx = -x/y
Then d/dx(x²+y²) = d/dx (25)
d/dx x² + d/dx y² = 0
2x + 2y dy/dx = 0
So we have 2x +2y (-3/4) = 0
Since dy/dx = -3/4 which is the slope at point (3,4) we have that the equation of the tangent line = 3x + 4y = 25