Sequence the steps of validation of content by placing proper step numbers.
a. Check whether the points included are correct or not.
b. Refer to various sources to know more about the topic.
c. Check spelling and grammar of the final slide.
d. Check whether all the relevant points are included in the presentation or not.
e. Add the missing points in the presentation.
f. Take notes in brief.
Answers
Explanation:
Topic :- This is your punishment for spamming in my question
Differentiation
To Differentiate :-
f(x)=\cot x \cdot \ln \sec xf(x)=cotx⋅lnsecx
Solution :-
f(x)=\cot x \cdot \ln \sec xf(x)=cotx⋅lnsecx
\dfrac{d(f(x))}{dx}=\dfrac{d(\cot x \cdot \ln \sec x)}{dx}
dx
d(f(x))
=
dx
d(cotx⋅lnsecx)
\dfrac{d(f(x))}{dx}=\ln \sec x\cdot\dfrac{d(\cot x )}{dx}+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}
dx
d(f(x))
=lnsecx⋅
dx
d(cotx)
+cotx⋅
dx
d(lnsecx)
\left(\because \dfrac{d(fg)}{dx}=g\cdot \dfrac{d(f)}{dx}+f\cdot\dfrac{d(g)}{dx} \right)(∵
dx
d(fg)
=g⋅
dx
d(f)
+f⋅
dx
d(g)
)
\dfrac{d(f(x))}{dx}=\ln \sec x\cdot(-\csc^2x)+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}
dx
d(f(x))
=lnsecx⋅(−csc
2
x)+cotx⋅
dx
d(lnsecx)
\left(\because \dfrac{d(\cot x)}{dx}=-\csc^2x \right)(∵
dx
d(cotx)
=−csc
2
x)
\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot\dfrac{d(\sec x )}{dx}
dx
d(f(x))
=−csc
2
x⋅lnsecx+cotx⋅
secx
1
⋅
dx
d(secx)
\left(\because \dfrac{d(\ln t)}{dx}=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)(∵
dx
d(lnt)
=
t
1
⋅
dx
dt
)
\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot \sec x\cdot \tan x
dx
d(f(x))
=−csc
2
x⋅lnsecx+cotx⋅
secx
1
⋅secx⋅tanx
\left(\because \dfrac{d(\sec x)}{dx}=\sec x\cdot \tan x \right)(∵
dx
d(secx)
=secx⋅tanx)
\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+(\cot x\cdot\tan x)\cdot\left (\dfrac{1}{\cancel{\sec x}}\cdot \cancel{\sec x}\right)
dx
d(f(x))
=−csc
2
x⋅lnsecx+(cotx⋅tanx)⋅(
secx
1
⋅
secx
)
\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+1
dx
d(f(x))
=−csc
2
x⋅lnsecx+1
(\because \cot x \cdot \tan x = 1)(∵cotx⋅tanx=1)
Answer :-
\underline{\boxed{\dfrac{d(f(x))}{dx}=1-\csc^2x \cdot\ln \sec x}}
dx
d(f(x))
=1−csc
2
x⋅lnsecx
Note : csc x = cosec x
\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}
ANSWER
Correct Expression :-
\mathtt{\dfrac{x^3+12x}{6x^2+8}=\dfrac{y^3+27y}{9y^2+27}}
6x
2
+8
x
3
+12x
=
9y
2
+27
y
3
+27y
To Find :-
\mathtt{x:y\:\:by\:using\:Componendo\:and\:Dividendo.}x:ybyusingComponendoandDividendo.
Concept Used :-
\mathtt{If\:\dfrac{a}{b}=\dfrac{c}{d},then\:using\:Componendo\:and\:Dividendo}If
b
a
=
d
c
,thenusingComponendoandDividendo
\mathtt{\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}}
a−b
a+b
=
c−d
c+d
Solution :-
\mathtt{\dfrac{x^3+12x}{6x^2+8}=\dfrac{y^3+27y}{9y^2+27}}
6x
2
+8
x
3
+12x
=
9y
2
+27
y
3
+27y
Using Componendo and Dividendo,
\mathtt{\dfrac{x^3+12x+(6x^2+8)}{x^3+12x-(6x^2+8)}=\dfrac{y^3+27y+(9y^2+27)}{y^3+27y-(9y^2+27)}}
x
3
+12x−(6x
2
+8)
x
3
+12x+(6x
2
+8)
=
y
3
+27y−(9y
2
+27)
y
3
+27y+(9y
2
+27)
Opening brackets,
\mathtt{\dfrac{x^3+12x+6x^2+8}{x^3+12x-6x^2-8}=\dfrac{y^3+27y+9y^2+27}{y^3+27y-9y^2-27}}
x
3
+12x−6x
2
−8
x
3
+12x+6x
2
+8
=
y
3
+27y−9y
2
−27
y
3
+27y+9y
2
+27
Rewriting it,
\mathtt{\dfrac{x^3+3(2)^2(x)+3(2)(x^2)+2^3}{x^3+3(2^2)x-3(2)(x^2)-2^3}=\dfrac{y^3+3(3^2)(y)+3(3)(y^2)+3^3}{y^3+3(3^2)(y)-3(3)(y^2)-3^3}}
x
3
+3(2
2
)x−3(2)(x
2
)−2
3
x
3
+3(2)
2
(x)+3(2)(x
2
)+2
3
=
y
3
+3(3
2
)(y)−3(3)(y
2
)−3
3
y
3
+3(3
2
)(y)+3(3)(y
2
)+3
3
We know that,
\mathtt{(a+b)^3=a^3+3ab^2+3a^2b+b^3}(a+b)
3
=a
3
+3ab
2
+3a
2
b+b
3
\mathtt{(a-b)^3=a^3+3ab^2-3a^2b-b^3}(a−b)
3
=a
3
+3ab
2
−3a
2
b−b
3
Using Identities,
\mathtt{\dfrac{(x+2)^3}{(x-2)^3}=\dfrac{(y+3)^3}{(y-3)^3}}
(x−2)
3
(x+2)
3
=
(y−3)
3
(y+3)
3
\mathtt{\left(\dfrac{x+2}{x-2}\right)^3=\left(\dfrac{y+3}{y-3}\right)^3}(
x−2
x+2
)
3
=(
y−3
y+3
)
3
Taking Cube Root on both sides,
\mathtt{\sqrt[3]{\left(\dfrac{x+2}{x-2}\right)^3}=\sqrt[3]{\left(\dfrac{y+3}{y-3}\right)^3}}
3
(
x−2
x+2
)
3
=
3
(
y−3
y+3
)
3
\mathtt{\dfrac{x+2}{x-2}=\dfrac{y+3}{y-3}}
x−2
x+2
=
y−3
y+3
Using Componendo and Dividendo again,
\mathtt{\dfrac{x}{2}=\dfrac{y}{3}}
2
x
=
3
y
We can write it as,
\mathtt{\dfrac{x}{y}=\dfrac{2}{3}}
y
x
=
3
2
Answer :-
Hence, x : y is equivalent to 2 : 3.
\begin{gathered}\Huge{\textbf{\textsf{{\purple{Ans}}{\pink{wer}}{\color{pink}{:}}}}} \\ \end{gathered}
