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MCV4U Unit 5, Derivatives of Trigonometric Functions, Derivatives of…
MCV4U Unit 5
Derivatives of Trigonometric Functions
(sinx)' = cosx
ex: composite functions
ex: f(x) = sin(x^3)
f'(x) = (d(sin(x^3))/dx)(d(x^3)/dx)
= 2xcos(x^2)
(tanx)' = 1 + (tanx)^2
= (secx)^2
These could be useful for optimizing triangles with an unknown angle/side length
Therefore it is useful to also remember SOHCAHTOA and the cosine law, also possibly some identities
sinx = opp / hyp
cosx = adj / hyp
tanx = opp / adj
c^2 = (a^2 + b^2 - c^2) / 2ab
(cosx)' = -sinx
ex: power rule
ex: f(x) = (cosx)^3
f'(x) = 3(cosx)^2(d(cosx)/dx)
= 3(cosx)^2(-sinx)
= -3(cosx)^2(sinx)
(cotx)' = -(1 + (cotx)^2) = -(cscx)^2
(sec)' = (sinx)(cosx)^-2 = (tanx)(secx)
(cscx)' = -(sinx)^-2(cosx) = -(cotx)(cscx)
Derivatives of Exponential Functions
e^x
d(e^x)/dx = e^x
d(e^ax)/dx = ae^ax
Where a is a constant
d(e^(x^a))/dx = (e^(x^a))(ax^a-1)
Where a is a constant
Note: This is not (e^x)^a. So you can't apply the power rule
d(e^g(x))/dx = (e^g(x))(g'(x))
ex: f(x) = e^(3x^2)
f'(x) = (e^(3x^2))(d(3x^2)/dx)
= 6x(e^(3x^2))
e is just a constant: 2.718...
When n --> infinity
e = ((1+1/n)/n)^n
ex: Compound interest - as the number of terms (n) increases, the limit will approach e
1/e = ((n-1)/n)^n
ex: Always getting 9/10 questions on a test correct, therefore as more and more questions are done (n), the limit approach 1/e
b^g(x)
d(b^g(x))/dx = (lnb)(b^g(x))(g'(x))
ex: f(x) = 2^(3x^2)
f'(x) = (ln2)(2^(3x^2))(d(3x^2)/dx)
= 6x^2(ln2)(2^(3x^2))
ex: f(x) = 2^(2x^1/2)
f'(x) = (ln2)(2^(2x^1/2))(d(2x^1/2)/dx)
= (ln2)(2^(2x^1/2))(1/2)(2x^-1/2)(2)
= (ln2)(2^(2x^1/2))(2x^-1/2)
d(a^e)/dx = 0 (where a is a constant)
Because e is just a constant and the derivative of a constant is 0
ex: d(4^e)/dx = 0
As a general guide, the derivative of (something^x) would be the derivative of "x" multiplied by (something^x)