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General

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Question

$fraction numerator text d end text over denominator text dx end text end fraction open square brackets s i n open parentheses l o g open parentheses s i n e to the power of x to the power of 2 end exponent end exponent close parentheses close parentheses close square brackets$ =

$e^{x^{2}} . c o t (e^{x^{2}})$
$c o s [l o g (s i n e^{x^{2}})] e^{x^{2}} c o t (e^{x^{2}})$
$c o s [l o g (s i n e^{x^{2}})] 2 x e^{x^{2}} c o t (e^{x^{2}})$
$e^{x^{2}} c o t (e^{x^{2}})$

The correct answer is: $c o s open square brackets l o g open parentheses s i n e to the power of x to the power of 2 end exponent end exponent close parentheses close square brackets 2 x e to the power of x to the power of 2 end exponent end exponent c o t open parentheses e to the power of x to the power of 2 end exponent end exponent close parentheses$

$fraction numerator d over denominator d x end fraction open curly brackets S i n open parentheses l o g open parentheses s i n e to the power of x to the power of 2 end exponent end exponent close parentheses close parentheses close curly brackets equals C o s open parentheses l o g left parenthesis s i n e to the power of x to the power of 2 end exponent end exponent right parenthesis close parentheses$ . $fraction numerator d over denominator d x end fraction l o g left parenthesis s i n e to the power of x to the power of 2 end exponent end exponent right parenthesis$

Related Questions to study

If $square root of T a n y end root$ = $e to the power of c o s 2 x end exponent. s i n x comma blank t h e n blank fraction numerator text dy end text over denominator text dx end text end fraction$ =

If $square root of T a n y end root$ = $e to the power of c o s 2 x end exponent. s i n x comma blank t h e n blank fraction numerator text dy end text over denominator text dx end text end fraction$ =

The line $chi plus gamma equals 2$ is tangent to tthe curve $chi to the power of 2 end exponent equals 3 minus 2 gamma$ at its point

The line $chi plus gamma equals 2$ is tangent to tthe curve $chi to the power of 2 end exponent equals 3 minus 2 gamma$ at its point

In general calcium carbonate is not soluble in water but by the addition of carbon dioxide to it, it becomes soluble. This is because:

In general calcium carbonate is not soluble in water but by the addition of carbon dioxide to it, it becomes soluble. This is because:

parallel

The correct stability order for the following species is:
I)
II)
III)
IV)

The correct stability order for the following species is:
I)
II)
III)
IV)

Chemistry-General

The point (4,3) is translated to the point (3,1) and then axes rotated through 30º about the origin then the new position of point is

The point (4,3) is translated to the point (3,1) and then axes rotated through 30º about the origin then the new position of point is

The arrangement of the following in ascending order of angle of rotation to eliminate xy term in the following equations.
i) 2 $text x end text to the power of 2 end exponent$ +5xy+2 $text y end text to the power of 2 end exponent$ +5x+7y+1=0
ii) 2 $text x end text to the power of 2 end exponent$ + $square root of text 3 end text end root$ xy+3 $text y end text to the power of 2 end exponent$ = 9
iii) 9 $x to the power of 2 end exponent$ +2 $square root of text 3 end text end root$ xy+7 $text y end text to the power of 2 end exponent$ = 10

The arrangement of the following in ascending order of angle of rotation to eliminate xy term in the following equations.
i) 2 $text x end text to the power of 2 end exponent$ +5xy+2 $text y end text to the power of 2 end exponent$ +5x+7y+1=0
ii) 2 $text x end text to the power of 2 end exponent$ + $square root of text 3 end text end root$ xy+3 $text y end text to the power of 2 end exponent$ = 9
iii) 9 $x to the power of 2 end exponent$ +2 $square root of text 3 end text end root$ xy+7 $text y end text to the power of 2 end exponent$ = 10

parallel

If the distance between the two given points is 2 units and the points are transferred by shifting the origin to (2,2) then the distance between the points in their new position is

If the distance between the two given points is 2 units and the points are transferred by shifting the origin to (2,2) then the distance between the points in their new position is

The transformed equation of $7 x to the power of 2 end exponent plus 2 square root of 3 x y plus 9 y to the power of 2 end exponent equals 8$ when the axes are rotated through an angle $fraction numerator text π end text over denominator text 3 end text end fraction$ is

The transformed equation of $7 x to the power of 2 end exponent plus 2 square root of 3 x y plus 9 y to the power of 2 end exponent equals 8$ when the axes are rotated through an angle $fraction numerator text π end text over denominator text 3 end text end fraction$ is

The transformed equation of $x to the power of 2 end exponent minus y to the power of 2 end exponent plus 2 x plus 4 y equals 0$ when the origin is shifted to the point (-1,2) is

The transformed equation of $x to the power of 2 end exponent minus y to the power of 2 end exponent plus 2 x plus 4 y equals 0$ when the origin is shifted to the point (-1,2) is

parallel

The point to which the origin should be shifted in order to eliminate x and y terms in the equation $4 x to the power of 2 end exponent plus 9 y to the power of 2 end exponent minus 8 x plus 36 y plus 4$ = 0

The point to which the origin should be shifted in order to eliminate x and y terms in the equation $4 x to the power of 2 end exponent plus 9 y to the power of 2 end exponent minus 8 x plus 36 y plus 4$ = 0

Orthocentre of the triangle is (2,1) and circumcentre is $open parentheses fraction numerator text 7 end text over denominator text 2 end text end fraction text , end text fraction numerator text 5 end text over denominator text 2 end text end fraction close parentheses$ , then its nine point circle centre is

Orthocentre of the triangle is (2,1) and circumcentre is $open parentheses fraction numerator text 7 end text over denominator text 2 end text end fraction text , end text fraction numerator text 5 end text over denominator text 2 end text end fraction close parentheses$ , then its nine point circle centre is

If a vertex of a $capital delta to the power of l e end exponent$ is (1,1) and the midpoints of two sides through this vertex are (-1,2)(3,2) then centroid of the triangle is

If a vertex of a $capital delta to the power of l e end exponent$ is (1,1) and the midpoints of two sides through this vertex are (-1,2)(3,2) then centroid of the triangle is

parallel

$I f blank A equals open parentheses 1 comma negative 1 close parentheses comma B equals open parentheses 2 comma 1 close parentheses comma C equals open parentheses 3 comma 0 close parentheses comma blank t h e n blank t h e blank c i r c u m minus r a d i u s blank o f blank increment A B C blank i s$

$I f blank A equals open parentheses 1 comma negative 1 close parentheses comma B equals open parentheses 2 comma 1 close parentheses comma C equals open parentheses 3 comma 0 close parentheses comma blank t h e n blank t h e blank c i r c u m minus r a d i u s blank o f blank increment A B C blank i s$

$T h e blank p o i n t s blank open parentheses a comma 0 close parentheses open parentheses 0 comma b close parentheses open parentheses 1 comma 1 close parentheses blank a r e blank c o l l i n e a r blank i f blank$

$T h e blank p o i n t s blank open parentheses a comma 0 close parentheses open parentheses 0 comma b close parentheses open parentheses 1 comma 1 close parentheses blank a r e blank c o l l i n e a r blank i f blank$

$I f fraction numerator pi over denominator 2 end fraction less than alpha less than pi comma blank t h e n blank t h e blank d i s t a n c e blank b e t w e e n blank t h e blank p o i n t s blank open parentheses t a n blank alpha comma 2 close parentheses comma$ $open parentheses 0 comma 1 close parentheses blank i s$

$I f fraction numerator pi over denominator 2 end fraction less than alpha less than pi comma blank t h e n blank t h e blank d i s t a n c e blank b e t w e e n blank t h e blank p o i n t s blank open parentheses t a n blank alpha comma 2 close parentheses comma$ $open parentheses 0 comma 1 close parentheses blank i s$

parallel

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