Trigonometry in Calculus: Limits, Slopes and Derivatives

Start with the small-angle limit

Class 11 foundation. When a small angle $h$ is measured in radians, its sine is close to the angle itself. This is a statement about a ratio approaching a value, not an exact equality at a nonzero angle:

The expression is undefined at $h=0$, because it would require division by zero. A limit asks what happens at nearby allowed inputs. Try $h=0.1$, then $0.01$: the ratio gets closer to $1$.

There is a geometric reason. In a unit circle, for $0<h<\pi/2$, comparing the inscribed triangle, sector and tangent triangle gives $\sin h<h<\tan h$. Dividing carefully yields $\cos h<\sin h/h<1$. Both outer values approach $1$, so the middle ratio must do the same. Negative inputs give the same ratio because both sine and the input change sign.

Radians make the unit-circle arc length equal to the angle. That is why this limit is $1$. If the number $d$ is a degree measure, then $\sin(d^\circ)=\sin(\pi d/180)$; the corresponding limit of $\sin(d^\circ)/d$ is $\pi/180$, not $1$.

Rescaling is often the whole solution. For a constant $a$, $\sin(ax)/x=a\,[\sin(ax)/(ax)]$, giving limit $a$. Another useful result follows from $1-\cos x=2\sin^2(x/2)$: $$(1-\cos x)/x^2\longrightarrow1/2.$$ Keep the original variable and the inner angle distinct while making the substitution.

From a shrinking secant to a tangent

Class 11 foundation. A secant joins two points on a graph. On $y=\sin x$, the points at $x$ and $x+h$ give slope $[\sin(x+h)-\sin x]/h$. As the second point approaches the first, the secant approaches the tangent when the derivative exists. The limit is the instantaneous slope (स्पर्शरेखा की ढाल).

Use the addition identity to uncover that limit:

The first quotient tends to $0$; the second tends to $1$. Consequently, $$\frac{d}{dx}\sin x=\cos x.$$ Expanding $\cos(x+h)$ in the same way gives $$\frac{d}{dx}\cos x=-\sin x.$$ These familiar rules are consequences of the angle-addition identities and radian limits.

Read the derivative as a number that changes with position. At $x=0$, sine has slope $1$. At $x=\pi/2$, its slope is $0$: the graph has a horizontal tangent at its peak. At $x=\pi$, its slope is $-1$. A zero derivative does not mean the function value is zero; here the peak value is $1$.

Move the two points in the slope explorer. Predict whether the tangent rises, falls or is horizontal before checking the value. Very small numerical steps may introduce rounding, so a displayed secant estimate need not equal the derivative exactly. The exact formula provides the target. Both trigonometric functions are differentiable for every real radian input; no gaps or poles occur in these two graphs.

Build the other four derivative rules

Class 11 foundation. Tangent, cotangent, secant and cosecant are built from sine and cosine. Their derivatives can therefore be derived with the quotient and reciprocal rules rather than memorised as unrelated facts.

For tangent, write $\tan x=\sin x/\cos x$, requiring $\cos x\neq0$. Applying the quotient rule gives $$(\tan x)'=\frac{\cos^2x+\sin^2x}{\cos^2x}=\sec^2x.$$ For cotangent, the numerator reverses: $(\cot x)'=-\operatorname{cosec}^2x$, where $\sin x\neq0$.

For example, differentiating $\sec x=(\cos x)^{-1}$ gives $\sin x/\cos^2x$, which equals $\sec x\tan x$. The two negative signs cancel. Differentiating $(\sin x)^{-1}$ instead leaves one negative sign, giving the cosecant rule. Check the sign by returning to the reciprocal.

A derivative rule applies on intervals where the original function is defined. Tangent does not suddenly acquire a derivative at $\pi/2$ because its formula looks compact; both the original function and its derivative expression fail there. On any continuous tangent branch, however, $\sec^2x>0$, so the graph is increasing. Cotangent has negative derivative on each of its continuous branches. These signs offer a quick visual check against the graphs you already know.

The reciprocal rules also explain what happens near a pole: a very small denominator can produce a large slope. That behaviour should agree with the steep branches of the graph, while the pole itself remains outside the domain.

Step into Class 12: follow the inner function

Class 12 route. A function such as $\sin(3x)$ contains two linked changes: $x$ changes the inner angle $3x$, and that angle changes its sine. The chain rule multiplies their rates:

Thus $(\sin(3x))'=3\cos(3x)$. Differentiating only the outside would miss how quickly the angle moves. Keep the inner expression intact until you have differentiated the outside.

For a differentiable inner function $u=u(x)$, the useful forms are $(\sin u)'=\cos u\,u'$, $(\cos u)'=-\sin u\,u'$, and $(\tan u)'=\sec^2u\,u'$. The last expression requires $\cos u\neq0$. The same pattern applies to the other reciprocal functions.

Distinguish a power from a changed angle. For $y=\sin^2x=(\sin x)^2$, the outside operation is squaring: $\frac{dy}{dx}=2\sin x\cos x$. For $y=\sin(x^2)$, the outside operation is sine: $\frac{dy}{dx}=2x\cos(x^2)$. These are different functions and usually have different slopes.

Combine rules in the order the expression demands. A product such as $x\cos(2x)$ needs the product rule, with the chain rule inside its second term. A quotient needs its quotient rule before simplifying. Before finishing, ask three questions: did you differentiate every changing factor, did you multiply by the inner derivative, and is the result being used only where the original function is defined? This small check catches many otherwise convincing errors.

At a specific point, differentiate first and then substitute its input. Substituting too early can turn a changing expression into a constant and erase the rate you wanted to find.

Differentiate a selected inverse angle

Class 12 route. Set $y=\sin^{-1}x$, so $\sin y=x$ and $y\in[-\pi/2,\pi/2]$. At an interior input $-1<x<1$, differentiate implicitly: $\cos y\,\frac{dy}{dx}=1$. Cosine is positive inside this selected branch, so $\cos y=\sqrt{1-x^2}$. Therefore $$\frac{d}{dx}\sin^{-1}x=\frac1{\sqrt{1-x^2}}\quad(-1<x<1).$$

The function itself is defined at $x=\pm1$, but this finite derivative formula is not. A function’s domain and its differentiability interval need not be identical.

For inverse tangent, differentiating $\tan y=x$ gives $\sec^2y\,\frac{dy}{dx}=1$. Since $\sec^2y=1+x^2$, its derivative is $1/(1+x^2)$. The negative inverse-cosine and inverse-cotangent signs agree with their decreasing graphs.

The absolute value in the inverse-secant and inverse-cosecant denominators matters for negative inputs. It comes from $\sqrt{x^2}=|x|$ when differentiating the reciprocal-input forms $\cos^{-1}(1/x)$ and $\sin^{-1}(1/x)$. Do not replace it by $x$ without restricting to positive inputs. For a composite inverse function, multiply by the inner derivative and apply the domain condition to that inner expression.

For inverse secant, the reciprocal-input derivative contains $\sqrt{1-1/x^2}=\sqrt{x^2-1}/|x|$. Combining it with the factor $1/x^2$ produces $1/(|x|\sqrt{x^2-1})$. This calculation explains both the absolute value and the requirement $|x|>1$, rather than leaving them as details to memorise.

Use the same ideas in a curve

Class 12 route. Trigonometry can appear in a relation, in a parametrised curve or in a second derivative. Decide what changes with the variable before choosing a rule.

In implicit differentiation, $y$ depends on $x$ even when it has not been isolated. From $\sin y=x$, differentiation gives $\cos y\,dy/dx=1$. From $x^2+\sin y=1$, it gives $$2x+\cos y\frac{dy}{dx}=0.$$ Thus $dy/dx=-2x/\cos y$, at points of the curve where $\cos y\neq0$. The chain-rule factor $dy/dx$ is essential; $y$ is not a constant.

For a parametrised curve $x=x(t),y=y(t)$, both coordinates change with a common parameter. Where $dx/dt\neq0$, $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}.$$ On $x=a\cos t,y=a\sin t$, with $a>0$, this becomes $-\cot t$ when $\sin t\neq0$. At a point where $dx/dt=0$, do not divide by zero: inspect the curve’s tangent separately.

A second derivative means differentiating the first derivative again. If $y=\sin(2x)$, then $\frac{dy}{dx}=2\cos(2x)$ and $\frac{d^2y}{dx^2}=-4\sin(2x)$. It is not $(\frac{dy}{dx})^2$. The second derivative describes how the slope changes; it helps interpret the bending of a graph.

These are different routes through the same chain of ideas. Choose one small curve, identify its input and output, write the required derivative relation, and check any denominator before simplifying. A clear diagram and a clear dependency statement often prevent more mistakes than a longer list of formulas.