Derivative of cos x with respect to x

Formula

$\dfrac{d}{dx} \, \cos x = -\sin x$

Proof

Cosine function is expressed as $\cos x$ trigonometrically if the angle of a right angled triangle is $x$. The derivative of $\cos x$ function can be derived in calculus with respect to $x$.

Step: 1

Use the fundamental expression of the derivative of a function with respect to $x$ in the form of the limit expression.

$\dfrac{d}{dx} \, f(x)$ $\,=\,$ $\displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

Take $f(x) = \cos x$ then $f(x+h) = \cos(x+h)$.

$\implies \dfrac{d}{dx} \, \cos x$ $\,=\,$ $\displaystyle \lim_{h \to 0} \dfrac{\cos (x+h) -\cos x}{h}$

Step: 2

The two cosine functions are involved in subtraction with different angles. The subtraction of them can be expressed in product form by applying sum to product conversion trigonometry identity.

$\cos \alpha -\cos \beta = -2 \sin \Bigg[\dfrac{\alpha+\beta}{2}\Bigg]\sin \Bigg[\dfrac{\alpha-\beta}{2}\Bigg]$

According to this identity, the numerator can be expressed in product form of sine functions and simplify the expression.

$\implies \dfrac{d}{dx} \, \cos x$ $\,=\,$ $\displaystyle \lim_{h \to 0} \dfrac{ -2\sin \Bigg[\dfrac{x+h+x}{2}\Bigg]\sin \Bigg[\dfrac{x+h-x}{2}\Bigg]}{h}$

$\implies \dfrac{d}{dx} \, \cos x$ $\,=\,$ $\require{cancel} \displaystyle \lim_{h \to 0} \dfrac{ -2\sin \Bigg[\dfrac{x+h+x}{2}\Bigg]\sin \Bigg[\dfrac{\cancel{x}+h-\cancel{x}}{2}\Bigg]}{h}$

$\implies \dfrac{d}{dx} \, \cos x$ $\,=\,$ $\displaystyle \lim_{h \to 0} \dfrac{ -2\sin \Bigg[\dfrac{2x+h}{2}\Bigg]\sin \Bigg[\dfrac{h}{2}\Bigg]}{h}$

Step: 3

The angle of the second sine term in numerator is $\dfrac{h}{2}$. The ratio contains $h$ in denominator and number $2$ in numerator. Use them to write them as $\dfrac{h}{2}$ in denominator.

$\implies \dfrac{d}{dx} \, \cos x$ $\,=\,$ $\displaystyle \lim_{h \to 0} \dfrac{ -\sin \Bigg[\dfrac{2x+h}{2}\Bigg]\sin \Bigg[\dfrac{h}{2}\Bigg]}{\Bigg[\dfrac{h}{2}\Bigg]}$

Now, split the numerator as two multiplying factors but keep $\dfrac{h}{2}$ as the denominator of the function $\sin \Bigg[\dfrac{h}{2}\Bigg]$.

$\implies \dfrac{d}{dx} \, \cos x$ $\,=\,$ $\displaystyle \lim_{h \to 0} -\sin \Bigg[\dfrac{2x+h}{2}\Bigg] \times \dfrac{\sin \Bigg[\dfrac{h}{2}\Bigg]}{\Bigg[\dfrac{h}{2}\Bigg]}$

Step: 4

The condition limit $h$ tends to zero is belong to both multiplying factors.

$\implies \dfrac{d}{dx} \, \cos x$ $\,=\,$ $\displaystyle \lim_{h \to 0} -\sin \Bigg[\dfrac{2x+h}{2}\Bigg]$ $\times$ $\displaystyle \lim_{h \to 0} \dfrac{\sin \Bigg[\dfrac{h}{2}\Bigg]}{\Bigg[\dfrac{h}{2}\Bigg]}$

No need to do any adjustments to the first multiplying factor but does for second multiplying factor. The function in numerator and denominator contain same. So, set the limit to same form to apply the limit h tends to zero sinx/x property.

If $h \to 0$ then $\dfrac{h}{2} \to \dfrac{0}{2}$. Therefore, $\dfrac{h}{2} \to 0$. Hence, $h \to 0$ can be replaced by $\dfrac{h}{2} \to 0$ in second multiplying factor.

$\implies \dfrac{d}{dx} \, \cos x$ $\,=\,$ $\displaystyle \lim_{h \to 0} -\sin \Bigg[\dfrac{2x+h}{2}\Bigg]$ $\times$ $\displaystyle \lim_{\frac{h}{2} \to 0} \dfrac{\sin \Bigg[\dfrac{h}{2}\Bigg]}{\Bigg[\dfrac{h}{2}\Bigg]}$

Substitute $h=0$ in the first multiplying factor and apply limit $h$ tends to zero $\dfrac{\sin x}{x}$ rule in the second multiplying factor.

$\implies \dfrac{d}{dx} \, \cos x$ $\,=\,$ $-\sin \Bigg[\dfrac{2x+0}{2}\Bigg]$ $\times 1$

$\implies \dfrac{d}{dx} \, \cos x$ $\,=\,$ $-\sin \Bigg[\dfrac{2x}{2}\Bigg] \times 1$

$\implies \dfrac{d}{dx} \, \cos x$ $\,=\,$ $\require{cancel} -\sin \Bigg[\dfrac{\cancel{2}x}{\cancel{2}}\Bigg]$

$\therefore \,\,\,\,\,\, \dfrac{d}{dx} \, \cos x \,=\, -\sin x$

Therefore, it is proved that the derivative of cosine of angle $x$ with respect to $x$ is equal to minus of sine of angle $x$. This property is used as a fundamental law in differentation to express derivative of cosine functions as sine functions.

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