Exercises — Derivatives of eˣ and aˣ — proofs

L1.1 Base $e$ hai, exponent exactly $x$ hai. Yeh "selfie" case hai: $f^{\prime }(x)=e^x.$

L1.2 Base $a=7$ (ek fixed number), exponent $x$ (variable). $\frac{d}{dx}{a}^x=a^x\ln a$ use karo $a=7$ ke saath: $g^{\prime }(x)=7^x\ln 7.$

L1.3 Sum ka derivative, derivatives ka sum hota hai. Pehla term $e^x$ apna khud ka derivative hai. Doosre term $x^5$ mein fixed exponent hai aur varying base $x$ hai, isliye yeh Power Rule case hai, $\frac{d}{dx}{x}^n=n{x}^{n-1}$: $h^{\prime }(x)=e^x+5x^4.$ Notice karo ki dono rules ek hi line mein hain: ek term mein exponent move kar raha hai ($e^x$), doosre mein base move kar raha hai ($x^5$).

Chain Rule kehta hai: agar $y=e^u$ jahan $u$ khud $x$ ka function hai, toh $\frac{dy}{dx}=e^u\cdot \frac{du}{dx}$. Yeh tool kyon? $e^x$ apna khud ka derivative tab hi hota hai jab exponent plain $x$ ho. Jab exponent koi bada expression $u$ ho, toh hume extra factor $\frac{du}{dx}$ — yaani "inner derivative" — dena padta hai.

L2.1 $u=5x$, toh $\frac{du}{dx}=5$: $y^{\prime }=e^{5x}\cdot 5=5e^{5x}.$

L2.2 $u=-3x$, toh $\frac{du}{dx}=-3$: $y^{\prime }=e^{-3x}\cdot (-3)=-3e^{-3x}.$ Negative inner derivative se slope negative hoti hai — yeh decay hai.

L2.3 Constant $4$ bas saath chalta hai (constant-multiple rule). Base $2$, exponent $x$: $y^{\prime }=4\cdot 2^x\ln 2.$

L2.4 $u=x^2$, toh $\frac{du}{dx}=2x$: $y^{\prime }=e^{x^2}\cdot 2x=2x{e}^{x^2}.$

L3.1 Base ko $e$ mein convert karo taaki Chain Rule clean rahe: $2^{x^2}=e^{(\ln 2)x^2}.$ Yahan $u=(\ln 2)x^2$, toh $\frac{du}{dx}=(\ln 2)(2x)=2x\ln 2$. Phir $y^{\prime }=e^{(\ln 2)x^2}\cdot 2x\ln 2=2^{x^2}(2x)\ln 2.$

L3.2 $\frac{d}{dx}{3}^x=3^x\ln 3$. $x=0$ par $3^0=1$ hai, toh slope hai $\ln 3\approx \mathrm{1.0986.}$ Yeh $1$ se bada hai: $3^x$, $e^x$ se zyaada steep start hota hai (jiska slope-at-$0$ exactly $1$ hai), kyunki $3>e$.

L3.3 Product rule $\frac{d}{dx}(uv)=u^{\prime }v+u{v}^{\prime }$ use karo $u=x^2$ ($u^{\prime }=2x$) aur $v=e^x$ ($v^{\prime }=e^x$) ke saath: $y^{\prime }=2x{e}^x+x^2{e}^x=e^x(x^2+2x).$

L3.4 Quotient rule $\frac{d}{dx}\frac{u}{v}=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$ use karo $u=e^x$, $v=x$ ke saath: $y^{\prime }=\frac{e^x\cdot x-e^x\cdot 1}{x^2}=\frac{e^x(x-1)}{x^2}(x\ne 0).$ Domain ka dhyan rakho: $x=0$ par original function undefined hai, toh wahan koi slope nahi.

L4.1 Direct: exponent $3x$ hai, toh $\frac{d}{dx}{a}^u=a^u\ln a\cdot u^{\prime }$ use karo $a=5$, $u=3x$, $u^{\prime }=3$ ke saath: $y^{\prime }=5^{3x}\ln 5\cdot 3=3\ln 5\cdot 5^{3x}.$ Base $e$ ke zariye: $5^{3x}=e^{(3\ln 5)x}$, toh $u=(3\ln 5)x$, $u^{\prime }=3\ln 5$: $y^{\prime }=e^{(3\ln 5)x}\cdot 3\ln 5=3\ln 5\cdot 5^{3x}.✓$ Same answer — dono raaste agree karne chahiye kyunki $a=e^{\ln a}$ ek identity hai.

L4.2 (Figure dekho.) $(a,e^a)$ point par slope hai $y^{\prime }(a)=e^a$ (self-derivative). Tangent line hai $y-e^a=e^a(x-a).$ $y=0$ set karo yeh find karne ke liye ki yeh $x$-axis ko kahan cross karti hai: $-e^a=e^a(x-a)⟹x-a=-1⟹x=a-1.$ $e^a$ cancel ho jaata hai — yahi wajah hai ki "1 unit left" $a$ se independent hai. Yeh constant sub-tangent $e^x$ ka signature hai kyunki woh apna khud ka derivative hai.

L4.3 "Height ke barabar rate se badhna" ka matlab hai slope $=$ height: $a^x\ln a=a^x.$ Kyunki $a^x>0$ sabhi $x$ ke liye, isse divide karo: $\ln a=1⟹a=e.$ Toh sirf base $e$ ki slope har jagah height ke barabar hoti hai — yeh $e$ ki definition se match karta hai. Kisi bhi doosre $a$ ke liye, slope $\ln a$ times height hoti hai (ek fixed multiplier, kabhi $1$ nahi jab tak $a=e$ na ho). Dekho The number e — definitions.

L5.1 $f(x)=a^x$ par derivative ki definition se shuru karo: $f^{\prime }(x)={\lim }_{h\to 0}\frac{a^{x+h}-a^x}{h}.$ Exponent law $a^{x+h}=a^x{a}^h$ use karo: $={\lim }_{h\to 0}\frac{a^x{a}^h-a^x}{h}={\lim }_{h\to 0}\frac{a^x(a^h-1)}{h}.$ Factor $a^x$ $h$ par depend nahi karta, toh yeh limit ke bahar nikal aata hai: $=a^x{\lim }_{h\to 0}\frac{a^h-1}{h}=a^{x}M(a).$ $M(a)$ kya hai: $a=e^{\ln a}$ likho, toh $a^x=e^{(\ln a)x}$; chain rule deta hai $\frac{d}{dx}{a}^x=e^{(\ln a)x}\cdot \ln a=a^x\ln a$. Match karne par, $M(a)=\ln a$. Geometrically $M(a)$ $a^x$ ki slope hai $x=0$ par (jahan height $a^0=1$ hai).

L5.2 $x^x$ mein variable dono base aur exponent mein hai, toh na power rule fit hota hai na plain exponential rule. Base $e$ ke zariye rewrite karo: $x^x=e^{x\ln x}.$ $u=x\ln x$ lo. Product rule se, $\frac{du}{dx}=1\cdot \ln x+x\cdot \frac{1}{x}=\ln x+1$. Chain rule: $\frac{d}{dx}{x}^x=e^{x\ln x}(\ln x+1)=x^x(\ln x+1).$

L5.3 L2-style chain rule se, $\frac{d}{dx}{e}^{kx}=k{e}^{kx}$: har differentiation $k$ se multiply karta hai. Ise $n$ baar apply karne par har baar ek $k$ bahar aata hai: $\frac{d}{dx}{e}^{kx}=k{e}^{kx},\frac{d^2}{d{x}^2}{e}^{kx}=k^2{e}^{kx},\dots ,\frac{d^n}{d{x}^n}{e}^{kx}=k^n{e}^{kx}.$ (Formal induction: base case $n=1$ dikha diya; agar $\frac{d^n}{d{x}^n}{e}^{kx}=k^n{e}^{kx}$, toh ek baar aur differentiate karne par $k^n\cdot k{e}^{kx}=k^{n+1}{e}^{kx}$ milta hai. ✓)

$\frac{d}{dx}{e}^{-3x}$? ::: $-3e^{-3x}$ $x=0$ par $3^x$ ki slope? ::: $\ln 3\approx 1.0986$ $\frac{d}{dx}(x^2{e}^x)$? ::: $e^x(x^2+2x)$ $\frac{d}{dx}\frac{e^x}{x}$? ::: $\frac{e^x(x-1)}{x^2}$ $(a,e^a)$ par $e^x$ ki tangent $x$-axis ko kahan milti hai? ::: $x=a-1$ par $\frac{d}{dx}{x}^x$? ::: $x^x(\ln x+1)$ $\frac{d^n}{d{x}^n}{e}^{kx}$? ::: $k^n{e}^{kx}$