# Calculus/Derivatives of Trigonometric Functions

 ← Product and Quotient Rules Calculus Chain Rule → Derivatives of Trigonometric Functions

Sine, cosine, tangent, cosecant, secant, cotangent. These are functions that crop up continuously in mathematics and engineering and have a lot of practical applications. They also appear in more advanced mathematics, particularly when dealing with things such as line integrals with complex numbers and alternate representations of space like spherical and cylindrical coordinate systems.

We use the definition of the derivative, i.e.,

${\displaystyle f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}$ ,

to work these first two out.

Let us find the derivative of sin(x), using the above definition.

 ${\displaystyle f(x)=\sin(x)}$ ${\displaystyle f'(x)=\lim _{h\to 0}{\frac {\sin(x+h)-\sin(x)}{h}}}$ Definition of derivative ${\displaystyle =\lim _{h\to 0}{\frac {\cos(x)\sin(h)+\cos(h)\sin(x)-\sin(x)}{h}}}$ trigonometric identity ${\displaystyle =\lim _{h\to 0}{\frac {\cos(x)\sin(h)+(\cos(h)-1)\sin(x)}{h}}}$ factoring ${\displaystyle =\lim _{h\to 0}{\frac {\cos(x)\sin(h)}{h}}+\lim _{h\to 0}{\frac {(\cos(h)-1)\sin(x)}{h}}}$ separation of terms ${\displaystyle =\cos(x)\times 1+\sin(x)\times 0}$ application of limit ${\displaystyle =\cos(x)}$ solution

Now for the case of cos(x).

 ${\displaystyle f(x)=\cos(x)}$ ${\displaystyle f'(x)=\lim _{h\to 0}{\frac {\cos(x+h)-\cos(x)}{h}}}$ Definition of derivative ${\displaystyle =\lim _{h\to 0}{\frac {\cos(x)\cos(h)-\sin(h)\sin(x)-\cos(x)}{h}}}$ trigonometric identity ${\displaystyle =\lim _{h\to 0}{\frac {\cos(x)(\cos(h)-1)-\sin(x)\sin(h)}{h}}}$ factoring ${\displaystyle =\lim _{h\to 0}{\frac {\cos(x)(\cos(h)-1)}{h}}-\lim _{h\to 0}{\frac {\sin(x)\sin(h)}{h}}}$ separation of terms ${\displaystyle =\cos(x)\times 0-\sin(x)\times 1}$ application of limit ${\displaystyle =-\sin(x)}$ solution

Therefore we have established

 Derivative of Sine and Cosine${\displaystyle {\frac {d}{dx}}\sin(x)=\cos(x)}$${\displaystyle {\frac {d}{dx}}\cos(x)=-\sin(x)}$

To find the derivative of the tangent, we just remember that:

${\displaystyle \tan(x)={\frac {\sin(x)}{\cos(x)}}}$

which is a quotient. Applying the quotient rule, we get:

${\displaystyle {\frac {d}{dx}}\tan(x)={\frac {\cos ^{2}(x)+\sin ^{2}(x)}{\cos ^{2}(x)}}}$

Then, remembering that ${\displaystyle \cos ^{2}(x)+\sin ^{2}(x)=1}$ , we simplify:

 ${\displaystyle {\frac {\cos ^{2}(x)+\sin ^{2}(x)}{\cos ^{2}(x)}}}$ ${\displaystyle ={\frac {1}{\cos ^{2}(x)}}}$ ${\displaystyle =\sec ^{2}(x)}$

 Derivative of the Tangent${\displaystyle {\frac {d}{dx}}\tan(x)=\sec ^{2}(x)}$

For secants, we again apply the quotient rule.

${\displaystyle \sec(x)={\frac {1}{\cos(x)}}}$
{\displaystyle {\begin{aligned}{\frac {d}{dx}}\sec(x)&={\frac {d}{dx}}{\frac {1}{\cos(x)}}\\&={\frac {\cos(x){\frac {d1}{dx}}-1{\frac {d\cos(x)}{dx}}}{\cos(x)^{2}}}\\&={\frac {\cos(x)0-1(-\sin(x))}{\cos(x)^{2}}}\end{aligned}}}

Leaving us with:

${\displaystyle {\frac {d}{dx}}\sec(x)={\frac {\sin(x)}{\cos ^{2}(x)}}}$

Simplifying, we get:

 Derivative of the Secant${\displaystyle {\frac {d}{dx}}\sec(x)=\sec(x)\tan(x)}$

Using the same procedure on cosecants:

${\displaystyle \csc(x)={\frac {1}{\sin(x)}}}$

We get:

 Derivative of the Cosecant${\displaystyle {\frac {d}{dx}}\csc(x)=-\csc(x)\cot(x)}$

Using the same procedure for the cotangent that we used for the tangent, we get:

 Derivative of the Cotangent${\displaystyle {\frac {d}{dx}}\cot(x)=-\csc ^{2}(x)}$
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