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Question-96538




Question Number 96538 by Hamida last updated on 02/Jun/20
Answered by john santu last updated on 02/Jun/20
y′=((2.7−2.8)/((8x+7)^2 )) = −(2/((8x+7)^2 ))
$$\mathrm{y}'=\frac{\mathrm{2}.\mathrm{7}−\mathrm{2}.\mathrm{8}}{\left(\mathrm{8x}+\mathrm{7}\right)^{\mathrm{2}} }\:=\:−\frac{\mathrm{2}}{\left(\mathrm{8x}+\mathrm{7}\right)^{\mathrm{2}} } \\ $$
Commented by Hamida last updated on 02/Jun/20
how it is become 2.7−2.8
$${how}\:{it}\:{is}\:{become}\:\mathrm{2}.\mathrm{7}−\mathrm{2}.\mathrm{8} \\ $$
Commented by john santu last updated on 02/Jun/20
in generally if   y=((ax+b)/(cx+d)) ⇒ y′=((a.d−b.c)/((cx+d)^2 ))
$$\mathrm{in}\:\mathrm{generally}\:\mathrm{if}\: \\ $$$$\mathrm{y}=\frac{{ax}+{b}}{{cx}+{d}}\:\Rightarrow\:\mathrm{y}'=\frac{{a}.{d}−{b}.{c}}{\left({cx}+{d}\right)^{\mathrm{2}} } \\ $$
Commented by Hamida last updated on 02/Jun/20
Thank you
$${Thank}\:{you}\: \\ $$$$ \\ $$

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