When it comes to calculus, derivatives are a key concept, and understanding how to differentiate functions like sin(x) is super important for exams like the SAT, ACT, and even the GRE or MCAT. The derivative of sin(x) is a fundamental topic in trigonometry and calculus, and once you get the hang of it, it becomes a tool to solve problems in areas like trigonometry, complex numbers, and more!

In this article, we’ll break down how to differentiate sin(x) using a few different rules, such as the first principle of derivatives, chain rule, and the quotient rule. Whether you’re tackling tough problems for your AP exam or prepping for the PERT or Accuplacer, understanding these rules and their applications will help you breeze through those calculus questions with confidence. So, let’s dive in and learn how to differentiate sin(x) like a pro with plenty of solved examples along the way!

What is Sinx? 

The derivative of a function is a concept in mathematics of a real variable that measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). They are a part of differential calculus. There are various methods of differentiation.  The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point.  The derivatives are used to find solutions to differential equations.

Sine is the trigonometric function of an angle. We can easily find out the Derivatives of Trigonometric Functions. The sine of an acute angle is defined in the context of a right triangle: sine is the ratio of the length of the side that is opposite that angle to the length of the longest side of the triangle (the hypotenuse). Cosine can be defined as an infinite series, or as the solution of certain differential equations, allowing their extension to arbitrary positive and negative values and even to complex numbers. In trigonometry, the law of sines, sine law, sine formula, or sine rule is an equation relating the lengths of the sides of a triangle (any shape) to the sines of its angles.

What is Derivative of Sinx?

 In calculus, the derivative represents the rate of change of a function with respect to its input.The derivative of sin(x) is cos(x). In mathematical terms, this can be written as d/dx sin(x) = cos(x). The sine function is a periodic function that oscillates between -1 and 1, and its derivative, the cosine function, represents the rate of change of the sine function at any point. Geometrically, the derivative of sin(x) corresponds to the slope of the curve of the sine function at each point.

This means that the rate of change of the sine function at any point x is equal to the cosine of that same point. Geometrically, this can be interpreted as the slope of the tangent line to the sine curve at the point (x, sin(x)). For trigonometric functions like sin(x), the derivative can be found using the rules of differentiation, which involve applying the chain rule and the derivative of elementary functions.

Graph of the Derivative of Sinx

The graph of the derivative of sin(x), denoted as d/dx(sin(x)) or cos(x), provides information on the rate of change of the sine function. The graph of the derivative of sin(x) is the graph of the cosine function, which is a periodic function that oscillates between 1 and -1 as x varies.

The graph of the derivative of sin(x) shows how the slope of the sine curve varies as x changes. Since the cosine function has a period of 2π, the graph of the derivative of sin(x) repeats every 2π units along the x-axis. The cosine function is an even function, meaning that it is symmetric about the y-axis, so the graph of the derivative of sin(x) is also symmetric about the y-axis. 

Derivative of Sinx Formula

The formula for the derivative of sin(x) is represented mathematically as d/dx(sin(x)) = cos(x). 

This formula represents the rate of change of the sine function with respect to x. It states that the slope of sin(x) at any given point x is equal to the cosine of x. The derivative of sin(x) can be obtained using the basic rules of calculus.The formula for the derivative of sin(x) can be derived using the limit definition of the derivative, the power rule, and the chain rule of differentiation. 

We will now learn how to differentiate sinx by using various differentiation rules like the first principle of derivative, Chain Rule and Quotient Rule.

Derivative of Sinx Proof by the First Principle

Derivative by the first principle refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change, which is equal to:

\(\begin{matrix}
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=sinx\\
f(x+h)=sin(x+h)\\
f(x+h)–f(x)= sin(x+h) – sin(x) = sinxcosh + cosxsinh – sinx\\
= sinx(cosh-1) + cosxsinh\\
{f(x+h) – f(x)\over{h}}={ sinx(cosh-1) + cosxsinh\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} { sinx(cosh-1) + cosxsinh\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinx(cosh-1)\over{h}} + \lim _{h{\rightarrow}0} {cosxsinh\over{h}}\\
= sinx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} + cosx \lim _{h{\rightarrow}0} {sinh\over{h}}\\
\text{Put h = 0 in first limit}\\
sinx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} = sinx\times0 = 0\\
\text{Using L’ Hospitals Rule on Second Limit}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = cosx \lim _{h{\rightarrow}0} {{d\over{dh}}sinh\over{{d\over{dh}}h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = cosx \lim _{h{\rightarrow}0} {cosh\over{1}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = cosx \times1 = cosx\\
f’(x)={dy\over{dx}} = {d(sinx)\over{dx}} = cosx
\end{matrix}\)

Alternatively,

\(\begin{matrix}\
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=sinx\\
f(x+h)=sin(x+h)\\
f(x+h)–f(x)= sin(x+h) – sin(x) = {2cos({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {2cos({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}}){sin({x+h-x\over{2}})\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0}2cos({x+h+x\over{2}}){sin({x+h-x\over{2}})\over{{h\over{2}}}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}})\times1\\
{\because}\lim _{h{\rightarrow}0}{sin({h\over{2}})\over{{h\over{2}}}} = 1\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}}) = cosx\\
f’(x)={dy\over{dx}} = {d(sinx)\over{dx}} = cosx
\end{matrix}\)

Derivative of Sinx Proof by the Chain Rule

Chain Rule helps us differentiate composite functions with the number of functions that make up the composition determining how many differentiation steps are necessary. The chain rule states that

\(\begin{matrix}
{d\over{dx}}f(g(x)) = f'(g(x))⋅g'(x).\\
{d\over{dx}}f[g(h(x))] = f'(g(h(x)))⋅g'(h(x))h'(x)
\end{matrix}\)

g and h, make up the composite function f, you have to consider the derivatives g′ and h′ in differentiating f( x).

If f(x) = sinx , find f’(x)

\(\begin{matrix}
f(x) = sinx = cos({\pi\over{2}} -x )\\
\text{ Using chain rule, }\\
f’(x) = – sin({\pi\over{2}} -x) · {d\over{dx}} ({\pi\over{2}} -x)\\
= – sin({\pi\over{2}} -x) · (-1)\\
= sin({\pi\over{2}} -x)\\
= cosx
\end{matrix}\)

Derivative of Sinx Proof by the Quotient Rule

A special rule, the quotient rule, exists for differentiating quotients of two functions. Functions often come as quotients, by which we mean one function divided by another function. There is a formula we can use to differentiate a quotient – it is called the quotient rule.

If f and g are both differentiable, then:

\({d\over{dx}}({f(x)\over{g(x)}}) = {g(x){d\over{dx}}f(x) – f(x){d\over{dx}}g(x)\over{g(x)}}\)

To solve the derivative by chain rule one must know the LIATE rule.

LIATE Rule

The LIATE Rule is as follows

A rule of thumb has been proposed, consisting of choosing as u the function that comes first in the following list:

• L – logarithmic functions: \({ln(x),log _{b}(x) }\) etc.
• I – inverse trigonometric functions (including hyperbolic analogues): arctan(x), arcsec(x), arsinh(x), etc.
• A – algebraic functions: \(x^{2},3x^{50}\).,tc.
• T – trigonometric functions (including hyperbolic analogues): sin(x), tan(x), {sech}(x), etc.
• E – exponential functions: \(e^{x},19^{x}\), etc.

If f(x) = sinx , find f’(x)

\(\begin{matrix}
f(x) = sinx = {1\over{cosecx}}\\
\text{ Using quotient rule, }\\
f’(x) = {cosecx{d\over{dx}}(1) – 1.{d\over{dx}}cosecx\over{cosec^2x}}\\
= {cosecxcotx\over{cosec^2x}}\\
= {cotx\over{cosecx}}\\
= {{cosx\over{sinx}}\over{{1\over{sinx}}}}\\
= cosx
\end{matrix}\)

Nth derivative of Sinx

The nth derivative of sin(x), denoted as dⁿ/dxⁿ(sin(x)), is a function that represents the rate of change of the sine function with respect to its input x, taken n times. It is given by the formula d^n/dx^n(sin(x)) = sin(x + nπ/2), where n is a non-negative integer. 

It states that the nth derivative of sin(x) is equal to the sine of the sum of x and n times π/2. For example, the first derivative of sin(x) is cos(x), which corresponds to the sine function with argument x + π/2. The second derivative of sin(x) is -sin(x), which corresponds to the sine function with argument x + π. The third derivative of sin(x) is -cos(x), which corresponds to the sine function with argument x + 3π/2, and so on. The nth derivative of sin(x) is a periodic function with a period of 2π/n, and it oscillates between positive and negative values as x varies. 

Anti-Derivative of sin x

Anti-Derivative of sin x is nothing but integration of sinx. Anti-derivative means inverse process of differentiation. As we know now the derivative of sin x is cos x. Hence, the anti-derivative of sin x is – cos x + C.

Solved Examples on Derivative of sin x

Solved Example: Find the derivative of \(sin^2x\)

Solution:

We solve this by using the Generalized Power Rule. The general power rule is a special case of the chain rule. It is useful when finding the derivative of a function that is raised to the nth power.

\({d\over{dx}}f(x)=n.f(x)^{n-1}\times{f'(x)}\)

Here,

\(\begin{matrix}
f(x) = sin^2x\\
{d\over{dx}}f(x) = {d\over{dx}}sin^2x\\
= 2sinx\times{d\over{dx}}(sinx)\\
= 2sinx\times(cosx)\\
= 2sinxcosx\\
= sin2x
\end{matrix}\)

Solved Example: Find the derivative of \(\frac{\sin x+\cos x}{\sin x-\cos x}\)

Solution:

Let \(f(x)=\frac{\sin x+\cos x}{\sin x-\cos x}\), then by quotient rule,

\(\begin{matrix} f^{\prime}(x)=& \frac{\frac{d}{d x}(\sin x+\cos x) \cdot(\sin x-\cos x)-(\sin x+\cos x) \cdot \frac{d}{d x}(\sin x-\cos x)}{(\sin x-\cos x)^{2}} \\=& \frac{(\cos x-\sin x)(\sin x-\cos x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^{2}} \\=& \frac{-\sin x \cos x-\sin ^{2} x-\cos ^{2} x-\sin x \cos x}{(\sin x-\cos x)^{2}} \\=& \frac{-2 \sin ^{2} x-2 \cos ^{2} x}{(\sin x-\cos x)^{2}}=\frac{-2\left(\sin ^{2} x+\cos ^{2} x\right)}{(\sin x-\cos x)^{2}} \\=& \frac{-2}{(\sin x-\cos x)^{2}}\left[\because \sin ^{2} x+\cos ^{2} x=1\right] \end{matrix}\)

Solved Example: Determine the derivative of the function \(f(x)={5\over{6}}sin⁡(x)\).

Solution:

Let \(f(x)={5\over{6}}sin⁡(x)\), then by constant rule,

We factor out the constant, ⅚.

\(f'(x)={5\over{6}}{dy\over{dx}}sin⁡(x)\)

\(f'(x)={5\over{6}}cos⁡(x)\)

Conclusion

In conclusion, understanding how to differentiate functions like sin(x) is essential for tackling calculus problems in exams like the SAT, ACT, and beyond. By using differentiation rules like the first principle, chain rule, and quotient rule, you'll be able to solve a wide range of problems confidently. With practice and a solid grasp of these concepts, differentiating sin(x) will become second nature, helping you ace your exams!