Exercises — Derivatives of all six trig functions

L1.1 By the sum picture above, differentiate term by term: $f^{\prime }(x)=\cos x+(-\sin x)=\cos x-\sin x.$ Concretely (see figure s01): the slope wave of $\sin x$ is $\cos x$, and the slope wave of $\cos x$ is $-\sin x$ (the same wave flipped, because $\cos$ falls where $\sin$ rises). Add those two slope waves and you get $f^{\prime }$.

L1.2 By the vertical-stretch picture, the constant $4$ multiplies the slope: $g^{\prime }(x)=4{\sec }^{2}x.$

L1.3 Term by term, using ${\sec }^{\prime }=\sec x\tan x$ and ${\csc }^{\prime }=-\csc x\cot x$ (re-derived at the top): $\frac{d}{dx}(\sec x-\csc x)=\sec x\tan x-(-\csc x\cot x)=\sec x\tan x+\csc x\cot x.$ WHAT just happened: the minus in front of $\csc x$ met the built-in minus of ${\csc }^{\prime }$, and two negatives made a plus.

L2.1 — Product Rule $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$ with $u=x^3,v=\cos x$. Two sides of the rectangle grow: $u^{\prime }=3x^2,v^{\prime }=-\sin x.$ $f^{\prime }(x)=3x^2\cos x+x^3(-\sin x)=3x^2\cos x-x^3\sin x.$ Notice the second term is negative: as $x$ grows, $\cos x$ (from figure s01) is on its way down, so that side of the rectangle is shrinking — the minus is the picture, not just algebra.

L2.2 — Chain Rule. Outer function $\tan (\square )$ gives ${\sec }^2(\square )$; inner $\square =5x$ gives $5$: $g^{\prime }(x)={\sec }^2(5x)\cdot 5=5{\sec }^2(5x).$ WHY the extra $5$: the chain rule says "derivative of outside $\times$ derivative of inside." Forgetting the inside factor is the classic error.

L2.3 — Quotient Rule ${\left(\frac{u}{v}\right)}^{\prime }=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$ with $u=\sin x$ ($u^{\prime }=\cos x$), $v=x^2$ ($v^{\prime }=2x$). Using the slope-of-a-ratio picture: $\sin$ rising pushes the ratio up ($+\cos x\cdot x^2$), while $x^2$ growing pushes it down ($-\sin x\cdot 2x$): $h^{\prime }(x)=\frac{\cos x\cdot x^2-\sin x\cdot 2x}{(x^2{)}^2}=\frac{x^2\cos x-2x\sin x}{x^4}.$ Cancel one $x$ (valid for $x\ne 0$): $h^{\prime }(x)=\frac{x\cos x-2\sin x}{x^3}.$

L3.1 $y^{\prime }={\sec }^{2}x=\frac{1}{{\cos }^{2}x}$. At $x=\pi /4$, $\cos (\pi /4)=\frac{\sqrt{2}}{2}$, so ${\cos }^2(\pi /4)=\frac{1}{2}$: $y^{\prime }{|}_{\pi /4}=\frac{1}{1/2}=2.$ Interpretation: the graph of $\tan$ is climbing at slope $2$ there — twice as steep as a $45^{\circ }$ line. In figure s04 the orange dot sits at $(\pi /4,1)$ and the dashed violet line through it has slope $2$; as $x$ moves toward the dotted asymptote $x=\pi /2$, $\cos x\to 0$ so ${\sec }^{2}x\to \infty$ and the curve rockets upward.

L3.2 Horizontal tangent means slope $=0$. For $y=\sin x$, $y^{\prime }=\cos x$, so we solve $\cos x=0⟹x=\frac{\pi }{2},\frac{3\pi }{2}.$ These are exactly the crest and trough of the sine wave — where it momentarily stops rising or falling. Look again at figure s01: at $x=\pi /2$ the magenta wave is flat and the violet slope-wave crosses zero.

L3.3 Point: at $x=\pi /2$, $y=\cos (\pi /2)=0$, so the point is $\left(\frac{\pi }{2},0\right)$. Slope: $y^{\prime }=-\sin x$, so $y^{\prime }{|}_{\pi /2}=-\sin (\pi /2)=-1$. Tangent line through $\left(\frac{\pi }{2},0\right)$ with slope $-1$: $y-0=-1\left(x-\frac{\pi }{2}\right)⟹y=-x+\frac{\pi }{2}.$

L4.1 — Product Rule (the two-sided rectangle) on the outside, Chain Rule on the $\sec (3x)$ factor. Let $u=x^2$ ($u^{\prime }=2x$) and $v=\sec (3x)$. For $v^{\prime }$: $v^{\prime }=\sec (3x)\tan (3x)\cdot 3=3\sec (3x)\tan (3x).$ Now assemble $f^{\prime }=u^{\prime }v+u{v}^{\prime }$ (side-strip plus top-strip): $f^{\prime }(x)=2x\sec (3x)+x^2\cdot 3\sec (3x)\tan (3x)=2x\sec (3x)+3x^2\sec (3x)\tan (3x).$ Factor the shared $\sec (3x)$: $f^{\prime }(x)=\sec (3x)[2x+3x^2\tan (3x)].$

L4.2 — Quotient Rule with $u=1+\sin x$ ($u^{\prime }=\cos x$), $v=\cos x$ ($v^{\prime }=-\sin x$). By the slope-of-a-ratio picture, numerator $u$ rising adds $\cos x\cdot \cos x$ and denominator $v$ falling (note $v^{\prime }=-\sin x$) contributes $-(1+\sin x)(-\sin x)$: $g^{\prime }(x)=\frac{\cos x\cdot \cos x-(1+\sin x)(-\sin x)}{{\cos }^{2}x}=\frac{{\cos }^{2}x+\sin x+{\sin }^{2}x}{{\cos }^{2}x}.$ Use the Pythagorean identity sin^2 + cos^2 = 1: the numerator ${\cos }^{2}x+{\sin }^{2}x=1$, leaving $g^{\prime }(x)=\frac{1+\sin x}{{\cos }^{2}x}.$ WHY this is beautiful: the identity collapses two of the three terms into a clean $1$. Domain caveat: this result holds only where $\cos x\ne 0$ — i.e. off the vertical asymptotes $x=\frac{\pi }{2}+k\pi$ ($k$ any integer). At those points the original $g(x)$ itself blows up, so neither $g$ nor $g^{\prime }$ is defined there. Never quote a quotient derivative without excluding its denominator's zeros.

L4.3 — Chain Rule. Outer $\sin (\square )$ gives $\cos (\square )$; inner $\square =\cos x$ gives $-\sin x$: $y^{\prime }=\cos (\cos x)\cdot (-\sin x)=-\sin x\cos (\cos x).$ WATCH: $\cos (\cos x)$ means "cosine of the number $\cos x$" — do not simplify it to ${\cos }^{2}x$. They are different beasts.

L5.1 Velocity is the derivative of position (Chain Rule, inner $2t$ gives $2$): $v(t)=\frac{d}{dt}[3\cos (2t)]=3\cdot (-\sin (2t))\cdot 2=-6\sin (2t)\text{m/s}.$ Acceleration is the derivative of velocity (chain rule again, inner gives $2$): $a(t)=\frac{d}{dt}[-6\sin (2t)]=-6\cos (2t)\cdot 2=-12\cos (2t){\text{m/s}}^2.$ Now compare to $x(t)=3\cos (2t)$: multiply $x$ by $-4$: $-4x(t)=-12\cos (2t)=a(t).✓$ So $a=-4x$, which is $a=-{\omega }^{2}x$ with ${\omega }^2=4$, i.e. $\boxed{\omega =2}$ rad/s. WHY $\omega$ is the inner derivative (figure s05): the position $3\cos (2t)$ is the shadow (horizontal projection) of a point spinning on a circle of radius $3$. In one second the point sweeps through angle $2t$, so it spins at $2$ radians per second — that spin rate is $\omega$. Differentiating asks "how fast is the angle changing?", and by the chain rule that rate is the inner derivative $\frac{d}{dt}(2t)=2$. Two derivatives spin the phasor twice, pulling out $2^2=4={\omega }^2$. The physics and the chain rule are the same fact.

L5.2 Quotient Rule with $u=\cos x$ ($u^{\prime }=-\sin x$), $v=\sin x$ ($v^{\prime }=\cos x$): $\frac{d}{dx}\cot x=\frac{(-\sin x)(\sin x)-(\cos x)(\cos x)}{{\sin }^{2}x}=\frac{-{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x}.$ Factor out $-1$ and apply ${\sin }^{2}x+{\cos }^{2}x=1$: $=\frac{-({\sin }^{2}x+{\cos }^{2}x)}{{\sin }^{2}x}=\frac{-1}{{\sin }^{2}x}=-{\csc }^{2}x.$ Undefined where? $\cot x$ and its derivative both blow up where $\sin x=0$, i.e. $x=0,\pm \pi ,\pm 2\pi ,\dots$ — the denominator vanishes there, so no slope exists (vertical asymptotes).

L5.3 Point: at $x=\pi /3$, $\cos (\pi /3)=\frac{1}{2}$, so $\sec (\pi /3)=\frac{1}{1/2}=2$. Point is $\left(\frac{\pi }{3},2\right)$. Slope: $y^{\prime }=\sec x\tan x$. With $\sec (\pi /3)=2$ and $\tan (\pi /3)=\sqrt{3}$: $y^{\prime }{|}_{\pi /3}=2\sqrt{3}.$ Tangent line through $\left(\frac{\pi }{3},2\right)$ with slope $2\sqrt{3}$: $y-2=2\sqrt{3}\left(x-\frac{\pi }{3}\right).$

Cover the answers. Can you state, from memory: (a) why the "co" functions carry a minus, (b) where $\sin x$ is flat, and (c) why differentiating $\cos (2t)$ twice produces a factor $4$? (a) They are the swapped-and-negated partners of $\sin ,\tan ,\sec$. ::: "All Co's are Negative." (b) Where $\cos x=0$, i.e. $x=\frac{\pi }{2},\frac{3\pi }{2}$ (crest and trough). ::: Slope $=\cos x$. (c) Each derivative pulls out the inner derivative $2$ via the chain rule; $2\times 2=4$. ::: That $4={\omega }^2$.