Chapter 5: Continuity and Differentiability (Calculus)

At $x=0$: ${\lim }_{x\to 0}(5x-3)=-3=f(0)$.

At $x=-3$: ${\lim }_{x\to -3}(5x-3)=-18=f(-3)$.

At $x=5$: ${\lim }_{x\to 5}(5x-3)=22=f(5)$.

In each case the limit equals the value, so f is continuous at all three points.

Limit = value at each point, so f is continuous at x = 0, −3, 5.

$f(3)=2(9)-1=17$ and ${\lim }_{x\to 3}(2x^2-1)=17$. They are equal.

f is continuous at x = 3.

(a) A polynomial: for every real c, ${\lim }_{x\to c}(x-5)=c-5=f(c)$. Continuous.

(b) Domain is $x\ne 5$. For $c\ne 5$, ${\lim }_{x\to c}\frac{1}{x-5}=\frac{1}{c-5}=f(c)$. Continuous at every point of its domain.

(c) For $x\ne -5$, $f(x)=\frac{(x-5)(x+5)}{x+5}=x-5$; for every c in the domain, ${\lim }_{x\to c}f(x)=c-5=f(c)$. Continuous.

(d) $f(x)=5-x$ for $x<5$ and $x-5$ for $x\ge 5$. Each piece is continuous; at $x=5$: LHL $=0$, RHL $=0$, $f(5)=0$. Continuous everywhere.

All four functions are continuous.

${\lim }_{x\to n}{x}^n=n^n=f(n)$ (a polynomial's limit is its value). So f is continuous at $x=n$.

${\lim }_{x\to n}{x}^n=n^n=f(n)$, so continuous.

At x = 0: near 0, $f(x)=x$, so ${\lim }_{x\to 0}f(x)=0=f(0)$. Continuous.

At x = 1: LHL $={\lim }_{x\to 1^{-}}x=1$, RHL $={\lim }_{x\to 1^{+}}5=5$. LHL ≠ RHL, so not continuous.

At x = 2: near 2, $f(x)=5$, so ${\lim }_{x\to 2}f(x)=5=f(2)$. Continuous.

Continuous at x = 0 and x = 2; not continuous at x = 1.

Both pieces are polynomials, so the only doubtful point is $x=2$.

LHL $=2(2)+3=7=f(2)$; RHL $=2(2)-3=1$. LHL ≠ RHL.

Discontinuous only at x = 2.

At x = −3: $f(-3)=3+3=6$; LHL $=6$; RHL $=-2(-3)=6$. Continuous.

At x = 3: LHL $=-2(3)=-6$; RHL $=f(3)=6(3)+2=20$. Not equal.

All other points lie inside one polynomial piece, so f is continuous there.

Discontinuous only at x = 3.

For $x<0$, $f(x)=-1$; for $x>0$, $f(x)=1$. At $x=0$: LHL $=-1$, RHL $=1$, unequal.

Discontinuous only at x = 0.

For $x<0$, $\frac{x}{|x|}=\frac{x}{-x}=-1$. So $f(x)=-1$ for every real x: a constant function, which is continuous everywhere.

No point of discontinuity.

At $x=1$: LHL $=1+1=2$, RHL $=f(1)=1+1=2$. Equal, so continuous; elsewhere f is a polynomial.

No point of discontinuity.

At $x=2$: LHL $=f(2)=8-3=5$, RHL $=4+1=5$. Continuous; elsewhere polynomial.

No point of discontinuity.

At $x=1$: LHL $=f(1)=1-1=0$, RHL $=1$. Unequal.

Discontinuous only at x = 1.

At $x=1$: LHL $=f(1)=6$, RHL $=1-5=-4$. Unequal, so f is discontinuous at $x=1$ (and continuous everywhere else).

No; f is not continuous at x = 1.

At x = 1: LHL $=3=f(1)$, RHL $=4$. Discontinuous.

At x = 3: LHL $=4$, RHL $=f(3)=5$. Discontinuous.

At all other points of $[0,10]$, f is constant nearby, hence continuous.

Discontinuous at x = 1 and x = 3 only.

At x = 0: LHL $=0$, $f(0)=0$, RHL $=0$. Continuous.

At x = 1: LHL $=f(1)=0$, RHL $=4$. Discontinuous.

x = 1 is the only point of discontinuity.

At x = −1: LHL $=f(-1)=-2$, RHL $=2(-1)=-2$. Continuous.

At x = 1: LHL $=f(1)=2$, RHL $=2$. Continuous.

f is continuous everywhere.

Need LHL = RHL = f(3): $3a+1=3b+3\Rightarrow 3a-3b=2$.

$a=b+\frac{2}{3}$

At $x=0$: LHL $=f(0)=\lambda (0)=0$; RHL $=4(0)+1=1$. These are never equal, whatever λ is. So f is discontinuous at 0 for every λ.

At $x=1$: near 1, $f(x)=4x+1$, so ${\lim }_{x\to 1}f(x)=5=f(1)$. Continuous for every λ.

No value of λ makes f continuous at 0; f is continuous at x = 1 for every λ.

Let n be an integer. $g(n)=n-n=0$.

LHL: for $n-1<x<n$, $[x]=n-1$, so $g(x)=x-n+1\to 1$.

RHL: for $n<x<n+1$, $[x]=n$, so $g(x)=x-n\to 0$.

LHL ≠ RHL, so g is discontinuous at every integer.

LHL = 1, RHL = 0 at each integer, so g is discontinuous there.

$f(\pi )={\pi }^2-0+5={\pi }^2+5$, and ${\lim }_{x\to \pi }(x^2-\sin x+5)={\pi }^2+5$. Equal.

Yes, f is continuous at x = π.

sin x and cos x are continuous for all real x. The sum, difference and product of continuous functions are continuous.

(a), (b), (c) are all continuous everywhere.

cos x: for any c, ${\lim }_{h\to 0}\cos (c+h)=\lim (\cos c\cos h-\sin c\sin h)=\cos c$. Continuous for all real x.

cosec x = 1/sin x: a quotient of continuous functions, continuous wherever $\sin x\ne 0$, i.e. except $x=n\pi$.

sec x = 1/cos x: continuous except where $\cos x=0$, i.e. $x=(2n+1)\frac{\pi }{2}$.

cot x = cos x/sin x: continuous except $x=n\pi$.

cos: everywhere; cosec, cot: except x = nπ; sec: except x = (2n + 1)π/2 (n ∈ Z).

At $x=0$: LHL $={\lim }_{x\to 0^{-}}\frac{\sin x}{x}=1$; RHL $=f(0)=1$. Continuous. Elsewhere each piece is continuous.

No point of discontinuity.

For $x\ne 0$ f is a product of continuous functions. At $x=0$: $-1\le \sin \frac{1}{x}\le 1\Rightarrow -x^2\le x^2\sin \frac{1}{x}\le x^2$. Both bounds → 0, so the limit is 0 $=f(0)$.

Yes, f is continuous for all real x.

At $x=0$: ${\lim }_{x\to 0}(\sin x-\cos x)=0-1=-1=f(0)$. Elsewhere continuous.

f is continuous for all real x.

Put $x=\frac{\pi }{2}+h$: $\cos x=-\sin h$, $\pi -2x=-2h$.

$$\underset{x\to \pi /2}{\lim }\frac{k\cos x}{\pi -2x}=\underset{h\to 0}{\lim }\frac{-k\sin h}{-2h}=\frac{k}{2}$$

$\frac{k}{2}=3\Rightarrow k=6$.

k = 6

LHL $=f(2)=4k$, RHL $=3$. So $4k=3$.

$k=\frac{3}{4}$

$k\pi +1=\cos \pi =-1\Rightarrow k\pi =-2$.

$k=-\frac{2}{\pi }$

$5k+1=15-5=10\Rightarrow k=\frac{9}{5}$.

$k=\frac{9}{5}$

At $x=2$: $5=2a+b$. At $x=10$: $10a+b=21$.

Subtracting: $8a=16\Rightarrow a=2$, then $b=1$.

a = 2, b = 1

$f=g\circ h$ with $h(x)=x^2$ (polynomial) and $g(x)=\cos x$, both continuous everywhere. A composite of continuous functions is continuous.

cos(x²) = (cos ∘ x²) is a composite of continuous functions, hence continuous.

$f=g\circ h$ with $h(x)=\cos x$ and $g(x)=|x|$, both continuous. So f is continuous.

Composite of continuous functions, hence continuous.

$\sin |x|=g(h(x))$ with $h(x)=|x|$ and $g(x)=\sin x$, both continuous. So $\sin |x|$ is continuous.

Continuous everywhere.

$|x|$ and $|x+1|$ are both continuous everywhere, so their difference is continuous everywhere.

(Piecewise: $f(x)=1$ for $x<-1$, $-2x-1$ for $-1\le x<0$, $-1$ for $x\ge 0$; checking $x=-1$: both sides give 1; $x=0$: both sides give −1.)

There is no point of discontinuity.

$$\frac{d}{dx}\sin (x^2+5)=\cos (x^2+5)\cdot 2x$$

$2x\cos (x^2+5)$

$$-\sin (\sin x)\cdot \cos x$$

$-\cos x\sin (\sin x)$

$\cos (ax+b)\cdot a$

$a\cos (ax+b)$

$$\frac{d}{dx}\sec (\tan \sqrt{x})=\sec (\tan \sqrt{x})\tan (\tan \sqrt{x})\cdot {\sec }^2\sqrt{x}\cdot \frac{1}{2\sqrt{x}}$$

$\frac{\sec (\tan \sqrt{x})\tan (\tan \sqrt{x}){\sec }^2\sqrt{x}}{2\sqrt{x}}$

Quotient rule:

$$\frac{a\cos (ax+b)\cos (cx+d)-\sin (ax+b)\cdot (-c\sin (cx+d))}{{\cos }^2(cx+d)}$$

$$=a\cos (ax+b)\sec (cx+d)+c\sin (ax+b)\tan (cx+d)\sec (cx+d)$$

$a\cos (ax+b)\sec (cx+d)+c\sin (ax+b)\tan (cx+d)\sec (cx+d)$

Product rule:

$$\frac{d}{dx}(\cos x^3)\cdot {\sin }^2{x}^5+\cos x^3\cdot \frac{d}{dx}({\sin }^2{x}^5)=-3x^2\sin x^3{\sin }^2{x}^5+\cos x^3\cdot 2\sin x^5\cos x^5\cdot 5x^4$$

$10x^4\sin x^5\cos x^5\cos x^3-3x^2\sin x^3{\sin }^2{x}^5$

$$\frac{dy}{dx}=2\cdot \frac{1}{2\sqrt{\cot x^2}}\cdot (-{\mathrm{cosec}}^2{x}^2)\cdot 2x=\frac{-2x{\mathrm{cosec}}^2{x}^2}{\sqrt{\cot x^2}}$$

Simplify: $\frac{{\mathrm{cosec}}^2{x}^2}{\sqrt{\cot x^2}}=\frac{1}{{\sin }^2{x}^2}\cdot \sqrt{\frac{\sin x^2}{\cos x^2}}=\frac{1}{\sin x^2\sqrt{\sin x^2\cos x^2}}=\frac{\sqrt{2}}{\sin x^2\sqrt{\sin 2x^2}}$.

$\frac{-2\sqrt{2}x}{\sin x^2\sqrt{\sin 2x^2}}$

$-\sin \sqrt{x}\cdot \frac{1}{2\sqrt{x}}$

$-\frac{\sin \sqrt{x}}{2\sqrt{x}}$

$$\text{LHD}=\underset{h\to 0^{-}}{\lim }\frac{f(1+h)-f(1)}{h}=\underset{h\to 0^{-}}{\lim }\frac{|h|}{h}=\underset{h\to 0^{-}}{\lim }\frac{-h}{h}=-1$$

$$\text{RHD}=\underset{h\to 0^{+}}{\lim }\frac{|h|}{h}=1$$

LHD ≠ RHD, so f is not differentiable at $x=1$.

LHD = −1, RHD = 1, so not differentiable at x = 1.

At x = 1: $f(1)=1$.

$\text{LHD}={\lim }_{h\to 0^{-}}\frac{[1+h]-1}{h}={\lim }_{h\to 0^{-}}\frac{0-1}{h}={\lim }_{h\to 0^{-}}\frac{-1}{h}$, which is not finite (→ +∞).

$\text{RHD}={\lim }_{h\to 0^{+}}\frac{[1+h]-1}{h}=\frac{1-1}{h}=0$.

They are not equal, so f is not differentiable at 1.

At x = 2: similarly LHD $={\lim }_{h\to 0^{-}}\frac{1-2}{h}$ does not exist, RHD $=0$. Not differentiable at 2.

One-sided derivatives differ at x = 1 and x = 2, so [x] is not differentiable there.

$2+3\frac{dy}{dx}=\cos x$

$\frac{dy}{dx}=\frac{\cos x-2}{3}$

$2+3\frac{dy}{dx}=\cos y\frac{dy}{dx}\Rightarrow \frac{dy}{dx}(\cos y-3)=2$.

$\frac{dy}{dx}=\frac{2}{\cos y-3}$

$a+2by\frac{dy}{dx}=-\sin y\frac{dy}{dx}\Rightarrow \frac{dy}{dx}(2by+\sin y)=-a$.

$\frac{dy}{dx}=\frac{-a}{2by+\sin y}$

$y+x\frac{dy}{dx}+2y\frac{dy}{dx}={\sec }^{2}x+\frac{dy}{dx}\Rightarrow \frac{dy}{dx}(x+2y-1)={\sec }^{2}x-y$.

$\frac{dy}{dx}=\frac{{\sec }^{2}x-y}{x+2y-1}$

$2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}(x+2y)=-(2x+y)$.

$\frac{dy}{dx}=-\frac{2x+y}{x+2y}$

$3x^2+\left(2xy+x^2\frac{dy}{dx}\right)+\left(y^2+2xy\frac{dy}{dx}\right)+3y^2\frac{dy}{dx}=0$

$\frac{dy}{dx}(x^2+2xy+3y^2)=-(3x^2+2xy+y^2)$

$\frac{dy}{dx}=-\frac{3x^2+2xy+y^2}{x^2+2xy+3y^2}$

$2\sin y\cos y\frac{dy}{dx}-\sin xy\left(y+x\frac{dy}{dx}\right)=0$

$\frac{dy}{dx}(\sin 2y-x\sin xy)=y\sin xy$

$\frac{dy}{dx}=\frac{y\sin xy}{\sin 2y-x\sin xy}$

$2\sin x\cos x-2\cos y\sin y\frac{dy}{dx}=0\Rightarrow \sin 2x=\sin 2y\frac{dy}{dx}$.

$\frac{dy}{dx}=\frac{\sin 2x}{\sin 2y}$

Put $x=\tan \theta$ (for $-1\le x\le 1$, $\theta \in [-\pi /4,\pi /4]$): $\frac{2\tan \theta }{1+{\tan }^2\theta }=\sin 2\theta$, so $y={\sin }^{-1}(\sin 2\theta )=2\theta =2{\tan }^{-1}x$.

$\frac{dy}{dx}=\frac{2}{1+x^2}$

Put $x=\tan \theta$, $\theta \in (-\pi /6,\pi /6)$: $\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }=\tan 3\theta$ and $3\theta \in (-\pi /2,\pi /2)$, so $y=3\theta =3{\tan }^{-1}x$.

$\frac{dy}{dx}=\frac{3}{1+x^2}$

Put $x=\tan \theta$, $\theta \in (0,\pi /4)$: $\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\cos 2\theta$ with $2\theta \in (0,\pi /2)$. So $y=2\theta =2{\tan }^{-1}x$.

$\frac{dy}{dx}=\frac{2}{1+x^2}$

${\sin }^{-1}u=\frac{\pi }{2}-{\cos }^{-1}u$, so by Q11, $y=\frac{\pi }{2}-2{\tan }^{-1}x$.

$\frac{dy}{dx}=-\frac{2}{1+x^2}$

$y=\frac{\pi }{2}-{\sin }^{-1}\frac{2x}{1+x^2}=\frac{\pi }{2}-2{\tan }^{-1}x$ (by Q9).

$\frac{dy}{dx}=-\frac{2}{1+x^2}$

Put $x=\sin \theta$, $\theta \in (-\pi /4,\pi /4)$: $2\sin \theta \cos \theta =\sin 2\theta$ with $2\theta \in (-\pi /2,\pi /2)$. So $y=2\theta =2{\sin }^{-1}x$.

$\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}$

$y={\cos }^{-1}(2x^2-1)$. Put $x=\cos \theta$, $\theta \in (\pi /4,\pi /2)$: $2{\cos }^2\theta -1=\cos 2\theta$ with $2\theta \in (\pi /2,\pi )$. So $y=2\theta =2{\cos }^{-1}x$.

$\frac{dy}{dx}=-\frac{2}{\sqrt{1-x^2}}$

$$\frac{e^x\sin x-e^x\cos x}{{\sin }^{2}x}$$

$\frac{e^x(\sin x-\cos x)}{{\sin }^{2}x}$, x ≠ nπ

$e^{{\sin }^{-1}x}\cdot \frac{1}{\sqrt{1-x^2}}$

$\frac{e^{{\sin }^{-1}x}}{\sqrt{1-x^2}}$, x ∈ (−1, 1)

$3x^2{e}^{x^3}$

$$\cos ({\tan }^{-1}{e}^{-x})\cdot \frac{1}{1+e^{-2x}}\cdot (-e^{-x})$$

$-\frac{e^{-x}\cos ({\tan }^{-1}{e}^{-x})}{1+e^{-2x}}$

$$\frac{1}{\cos e^x}\cdot (-\sin e^x)\cdot e^x$$

$-e^x\tan e^x$, $e^x\ne (2n+1)\frac{\pi }{2}$

Differentiate each term by the chain rule: $\frac{d}{dx}{e}^{x^k}=k{x}^{k-1}{e}^{x^k}$.

$e^x+2x{e}^{x^2}+3x^2{e}^{x^3}+4x^3{e}^{x^4}+5x^4{e}^{x^5}$

$$\frac{1}{2\sqrt{e^{\sqrt{x}}}}\cdot e^{\sqrt{x}}\cdot \frac{1}{2\sqrt{x}}$$

$\frac{e^{\sqrt{x}}}{4\sqrt{x{e}^{\sqrt{x}}}}$

$\frac{1}{\log x}\cdot \frac{1}{x}$

$\frac{1}{x\log x}$

$$\frac{-\sin x\log x-\cos x\cdot \frac{1}{x}}{(\log x{)}^2}$$

$-\frac{x\sin x\log x+\cos x}{x(\log x{)}^2}$

$-\sin (\log x+e^x)\cdot \left(\frac{1}{x}+e^x\right)$

$-\left(\frac{1}{x}+e^x\right)\sin (\log x+e^x)$

$\log y=\log \cos x+\log \cos 2x+\log \cos 3x$

$\frac{1}{y}\frac{dy}{dx}=-\tan x-2\tan 2x-3\tan 3x$

$-\cos x\cos 2x\cos 3x[\tan x+2\tan 2x+3\tan 3x]$

$\log y=\frac{1}{2}[\log (x-1)+\log (x-2)-\log (x-3)-\log (x-4)-\log (x-5)]$

$\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]$

$\log y=\cos x\cdot \log (\log x)$

$\frac{1}{y}\frac{dy}{dx}=-\sin x\log (\log x)+\cos x\cdot \frac{1}{x\log x}$

$(\log x{)}^{\cos x}\left[\frac{\cos x}{x\log x}-\sin x\log (\log x)\right]$

Let $u=x^x$: $\log u=x\log x\Rightarrow \frac{du}{dx}=x^x(1+\log x)$.

$\frac{d}{dx}{2}^{\sin x}=2^{\sin x}\log 2\cdot \cos x$.

$x^x(1+\log x)-2^{\sin x}\cos x\log 2$

$\log y=2\log (x+3)+3\log (x+4)+4\log (x+5)$

$\frac{1}{y}\frac{dy}{dx}=\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}=\frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}$

Numerator $=2x^2+18x+40+3x^2+24x+45+4x^2+28x+48=9x^2+70x+133$.

$(x+3)(x+4{)}^2(x+5{)}^3(9x^2+70x+133)$

Let $u={\left(x+\frac{1}{x}\right)}^x$: $\log u=x\log \left(x+\frac{1}{x}\right)$,

$\frac{du}{dx}=u\left[\log \left(x+\frac{1}{x}\right)+x\cdot \frac{1-\frac{1}{x^2}}{x+\frac{1}{x}}\right]=u\left[\log \left(x+\frac{1}{x}\right)+\frac{x^2-1}{x^2+1}\right]$.

Let $v=x^{1+\frac{1}{x}}$: $\log v=\left(1+\frac{1}{x}\right)\log x$,

$\frac{dv}{dx}=v\left[-\frac{1}{x^2}\log x+\left(1+\frac{1}{x}\right)\frac{1}{x}\right]=v\cdot \frac{x+1-\log x}{x^2}$.

${\left(x+\frac{1}{x}\right)}^x\left[\frac{x^2-1}{x^2+1}+\log \left(x+\frac{1}{x}\right)\right]+x^{1+\frac{1}{x}}\left(\frac{x+1-\log x}{x^2}\right)$

$u=(\log x{)}^x$: $\log u=x\log (\log x)$, $\frac{du}{dx}=u\left[\log (\log x)+\frac{1}{\log x}\right]=(\log x{)}^{x-1}[1+\log x\log (\log x)]$.

$v=x^{\log x}$: $\log v=(\log x{)}^2$, $\frac{dv}{dx}=v\cdot \frac{2\log x}{x}=2x^{\log x-1}\log x$.

$(\log x{)}^{x-1}[1+\log x\cdot \log (\log x)]+2x^{\log x-1}\log x$

$u=(\sin x{)}^x$: $\log u=x\log \sin x$, $\frac{du}{dx}=(\sin x{)}^x(\log \sin x+x\cot x)$.

$\frac{d}{dx}{\sin }^{-1}\sqrt{x}=\frac{1}{\sqrt{1-x}}\cdot \frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x-x^2}}$.

$(\sin x{)}^x(x\cot x+\log \sin x)+\frac{1}{2\sqrt{x-x^2}}$

$u=x^{\sin x}$: $\log u=\sin x\log x$, $\frac{du}{dx}=x^{\sin x}\left(\cos x\log x+\frac{\sin x}{x}\right)$.

$v=(\sin x{)}^{\cos x}$: $\log v=\cos x\log \sin x$, $\frac{dv}{dx}=(\sin x{)}^{\cos x}(-\sin x\log \sin x+\cos x\cot x)$.

$x^{\sin x}\left(\frac{\sin x}{x}+\cos x\log x\right)+(\sin x{)}^{\cos x}(\cos x\cot x-\sin x\log \sin x)$

$u=x^{x\cos x}$: $\log u=x\cos x\log x$,

$\frac{du}{dx}=u[\cos x\log x-x\sin x\log x+\cos x]=x^{x\cos x}[\cos x(1+\log x)-x\sin x\log x]$.

$\frac{d}{dx}\frac{x^2+1}{x^2-1}=\frac{2x(x^2-1)-2x(x^2+1)}{(x^2-1{)}^2}=\frac{-4x}{(x^2-1{)}^2}$.

$x^{x\cos x}[\cos x(1+\log x)-x\sin x\log x]-\frac{4x}{(x^2-1{)}^2}$

$u=(x\cos x{)}^x$: $\log u=x\log (x\cos x)=x[\log x+\log \cos x]$,

$\frac{du}{dx}=u[\log (x\cos x)+1-x\tan x]$.

$v=(x\sin x{)}^{1/x}$: $\log v=\frac{1}{x}[\log x+\log \sin x]$,

$\frac{dv}{dx}=v\left[-\frac{1}{x^2}\log (x\sin x)+\frac{1}{x}\left(\frac{1}{x}+\cot x\right)\right]=v\cdot \frac{x\cot x+1-\log (x\sin x)}{x^2}$.

$(x\cos x{)}^x[1-x\tan x+\log (x\cos x)]+(x\sin x{)}^{\frac{1}{x}}\left[\frac{x\cot x+1-\log (x\sin x)}{x^2}\right]$

$\frac{d}{dx}{x}^y=x^y\left(\frac{dy}{dx}\log x+\frac{y}{x}\right)$ and $\frac{d}{dx}{y}^x=y^x\left(\log y+\frac{x}{y}\frac{dy}{dx}\right)$.

Sum $=0$: $\frac{dy}{dx}(x^y\log x+x{y}^{x-1})=-(y{x}^{y-1}+y^x\log y)$.

$\frac{dy}{dx}=-\frac{y{x}^{y-1}+y^x\log y}{x^y\log x+x{y}^{x-1}}$

$x\log y=y\log x$. Differentiate: $\log y+\frac{x}{y}\frac{dy}{dx}=\frac{dy}{dx}\log x+\frac{y}{x}$.

$\frac{dy}{dx}\left(\frac{x}{y}-\log x\right)=\frac{y}{x}-\log y$

$\frac{dy}{dx}=\frac{y}{x}\left(\frac{y-x\log y}{x-y\log x}\right)$

$y\log \cos x=x\log \cos y$. Differentiate:

$\frac{dy}{dx}\log \cos x-y\tan x=\log \cos y-x\tan y\frac{dy}{dx}$

$\frac{dy}{dx}(\log \cos x+x\tan y)=\log \cos y+y\tan x$

$\frac{dy}{dx}=\frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}$

$\log x+\log y=x-y$. Differentiate: $\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}$.

$\frac{dy}{dx}\left(\frac{1+y}{y}\right)=\frac{x-1}{x}$

$\frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}$

$\log f=\log (1+x)+\log (1+x^2)+\log (1+x^4)+\log (1+x^8)$

$$f^{\prime }(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)\left[\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}\right]$$

$f^{\prime }(1)=16\left[\frac{1}{2}+1+2+4\right]=16\times \frac{15}{2}=120$.

$f^{\prime }(1)=120$

(i) Product rule: $(2x-5)(x^3+7x+9)+(x^2-5x+8)(3x^2+7)$

$=(2x^4-5x^3+14x^2-17x-45)+(3x^4-15x^3+31x^2-35x+56)=5x^4-20x^3+45x^2-52x+11$.

(ii) Expanding: $y=x^5-5x^4+15x^3-26x^2+11x+72$, so $\frac{dy}{dx}=5x^4-20x^3+45x^2-52x+11$.

(iii) Log: $\frac{1}{y}\frac{dy}{dx}=\frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}$; multiplying by y gives exactly the expression in (i), $=5x^4-20x^3+45x^2-52x+11$.