using-the-1st-principle-find-the-derivative-of-y-ax-b-n-

Question Number 25845 by NECx last updated on 15/Dec/17

$${using}\:{the}\:\mathrm{1}{st}\:{principle}\:{find}\:{the} \\ $$$${derivative}\:{of}\: \\ $$$$\:\:\:\:\:{y}=\left({ax}+{b}\right)^{{n}} \\ $$

Answered by ajfour last updated on 15/Dec/17

(dy/dx)=lim_(h→0) (((ax+ah+b)^n −(ax+b)^n )/h) =lim_(h→0) ((ah)/h)×lim_(h→0) [(ax+ah+b)^(n−1) + (ax+ah+b)^(n−2) (ax+b)+ (ax+ah+b)^(n−3) (ax+b)^2 +... .....+ (ax+ah+b)(ax+b)^(n−2) + (ax+b)^(n−1) ] ((d(ax+b)^n )/dx)=an(ax+b)^(n−1) .

$$\frac{{dy}}{{dx}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({ax}+{ah}+{b}\right)^{{n}} −\left({ax}+{b}\right)^{{n}} }{{h}} \\ $$$$\:\:\:\:=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ah}}{{h}}×\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\left({ax}+{ah}+{b}\right)^{{n}−\mathrm{1}} +\right. \\ $$$$\:\:\:\:\:\:\left({ax}+{ah}+{b}\right)^{{n}−\mathrm{2}} \left({ax}+{b}\right)+ \\ $$$$\:\:\:\:\:\:\left({ax}+{ah}+{b}\right)^{{n}−\mathrm{3}} \left({ax}+{b}\right)^{\mathrm{2}} +… \\ $$$$\:\:\:…..+\:\left({ax}+{ah}+{b}\right)\left({ax}+{b}\right)^{{n}−\mathrm{2}} + \\ $$$$\left.\:\:\:\:\:\:\:\left({ax}+{b}\right)^{{n}−\mathrm{1}} \right] \\ $$$$\:\:\:\:\frac{{d}\left({ax}+{b}\right)^{{n}} }{{dx}}={an}\left({ax}+{b}\right)^{{n}−\mathrm{1}} \:. \\ $$

Commented by NECx last updated on 16/Dec/17

i still dont get how you removed lim_(x→0) ((ah)/h)

$${i}\:{still}\:{dont}\:{get}\:{how}\:{you}\:{removed} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{ah}}{{h}} \\ $$

Commented by ajfour last updated on 16/Dec/17

x^n −y^n =(x−y)(x^(n−1) +x^(n−2) y+.. ...+xy^(n−2) +y^(n−1) ) so in your question we have x^n =(ax+ah+b)^n and y^n =(ax+b)^n so (x−y)=ah.

$${x}^{{n}} −{y}^{{n}} =\left({x}−{y}\right)\left({x}^{{n}−\mathrm{1}} +{x}^{{n}−\mathrm{2}} {y}+..\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…+{xy}^{{n}−\mathrm{2}} +{y}^{{n}−\mathrm{1}} \right) \\ $$$${so}\:{in}\:{your}\:{question}\:{we}\:{have} \\ $$$$\:\:{x}^{{n}} =\left({ax}+{ah}+{b}\right)^{{n}} \\ $$$${and}\:\:\:{y}^{{n}} =\left({ax}+{b}\right)^{{n}} \\ $$$${so}\:\:\:\left({x}−{y}\right)={ah}. \\ $$

Answered by rita1608 last updated on 15/Dec/17

((lim)/(h→0)) (((a(x+h)+b)^n −(ax+b)^n )/h) =((lim)/(h→0)) (((ax+ah+b)^n −(ax+b)^n )/h) =((lim)/(h→0)) (((ax+b)^n −(ax+b)^n )/h)=0

$$\frac{{lim}}{{h}\rightarrow\mathrm{0}}\:\:\frac{\left({a}\left({x}+{h}\right)+{b}\right)^{{n}} −\left({ax}+{b}\right)^{{n}} }{{h}} \\ $$$$=\frac{{lim}}{{h}\rightarrow\mathrm{0}}\:\:\:\frac{\left({ax}+{ah}+{b}\right)^{{n}} −\left({ax}+{b}\right)^{{n}} }{{h}} \\ $$$$=\frac{{lim}}{{h}\rightarrow\mathrm{0}}\:\frac{\left({ax}+{b}\right)^{{n}} −\left({ax}+{b}\right)^{{n}} }{{h}}=\mathrm{0} \\ $$

Commented by rita1608 last updated on 16/Dec/17

$${thanks}\:{for}\:{the}\:{correction}\: \\ $$$$ \\ $$

Commented by prakash jain last updated on 15/Dec/17

In the last step you substituted h=0 in numerator and left h in denominator. This is not correct. with this you will get 0 for every derivative.

$$\mathrm{In}\:\mathrm{the}\:\mathrm{last}\:\mathrm{step}\:\mathrm{you}\:\mathrm{substituted} \\ $$$${h}=\mathrm{0}\:\mathrm{in}\:\mathrm{numerator}\:\mathrm{and}\:\mathrm{left} \\ $$$${h}\:\mathrm{in}\:\mathrm{denominator}.\:\mathrm{This}\:\mathrm{is}\:\mathrm{not}\:\mathrm{correct}. \\ $$$$\mathrm{with}\:\mathrm{this}\:\mathrm{you}\:\mathrm{will}\:\mathrm{get}\:\mathrm{0}\:\mathrm{for}\:\mathrm{every} \\ $$$$\mathrm{derivative}. \\ $$

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