Stewart – Calculus – 3.3 – Derivatives of Trigonometric Functions

$\frac{d}{\mathit{dx}}(\sin x)=?$ $\frac{d}{\mathit{dx}}(\cos x)=?$ $\frac{d}{\mathit{dx}}(\tan x)=?$ $\frac{d}{\mathit{dx}}(\mathrm{csc} x)=?$ $\frac{d}{\mathit{dx}}(\mathrm{sec} x)=?$ $\frac{d}{\mathit{dx}}(\mathrm{cot} x)=?$

THE TRICK

To learn the derivatives of the 6 trig functions you actually only have to learn 3 of them. The three to learn are sine, tangent, and secant.

IF	THEN
F(X) = sin X	F'(X) = cos X
F(X) = tan X	F'(X) = sec2 X
F(X) = sec X	F'(X) = sec X tan X

The other three functions all start with "co":
Cosine, Cotangent, and Cosecant.
Think of the functions as having partners:

sine	to	cosine
tangent	to	cotangent
secant	to	cosecant

To find the derivative of the "co" functions, start with the derivatives of sine, tangent and secant, change each function in the derivative to it’s co-function partner and put a minus sign in front of the derivative.

So starting with one of the three we need to know:

If F(x) = secant X
then F'(x) = secant X tangent X

So now to get the derivative of the cosecant:
If F(X) = cosecant X
then F'(X) = -cosecant X cotangent X

So all 6 trig function derivatives look like this:

IF:	THEN:
F(X) = sinX	F'(X) = cosX
F(X) = cosX	F'(X) = – sinX
F(X) = tanX	F'(X) = sec2 X
F(X) = cotX	F'(X) = – csc2 X
F(X) = secX	F'(X) = sec X tan X
F(X) = cscX	F'(X) = -csc X cot X

Thanks to Bruce Kirkpatrick for this.

planet & epoch

Stewart – Calculus – 3.3 – Derivatives of Trigonometric Functions

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