Solution for the question - ify=tan^(-1)((sqrt(1+a^(2)x^(2))-1)/(ax))then(1+a^(2)x^(2))y^(||)+2a^(2)xy^(-)= 0-2 a^(2) '/> 2 a^(2) '/>

For such questions, we should know the formulas of inverse functions.

If $y equals s i n invisible function application 2 x s i n invisible function application 3 x s i n invisible function application 4 x text then end text y to the power of parallel to end exponent equals$

For such questions, we should know different formulas.

If $y equals s i n invisible function application 2 x s i n invisible function application 3 x s i n invisible function application 4 x text then end text y to the power of parallel to end exponent equals$

For such questions, we should know different formulas.

$fraction numerator d over denominator d x end fraction open curly brackets Tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator square root of 1 plus x to the power of 2 end exponent end root minus 1 over denominator x end fraction close parentheses close curly brackets equals$

For such questions, we will use different formulas.

$fraction numerator d over denominator d x end fraction open curly brackets Tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator square root of 1 plus x to the power of 2 end exponent end root minus 1 over denominator x end fraction close parentheses close curly brackets equals$

For such questions, we will use different formulas.

The length of the perpendicular from the incentre of the triangle formed by the axes and the line $fraction numerator x over denominator 3 end fraction plus fraction numerator y over denominator 4 end fraction equals 1$ to the hypotenuse is

If $theta$ is the angle between two adjacent sides of a parallelogram and p, q are the distances between the parallel sides, then the area of the parallelogram

$fraction numerator d over denominator d x end fraction open parentheses fraction numerator a x plus b over denominator c x plus d end fraction close parentheses equals$

For such questions, we should know how to use u by v method.

$fraction numerator d over denominator d x end fraction open parentheses fraction numerator a x plus b over denominator c x plus d end fraction close parentheses equals$

For such questions, we should know how to use u by v method.

$fraction numerator d over denominator d x end fraction open parentheses e to the power of x to the power of 2 end exponent end exponent a to the power of x end exponent close parentheses equals$

For such questions, we should know u.v method.

$fraction numerator d over denominator d x end fraction open parentheses e to the power of x to the power of 2 end exponent end exponent a to the power of x end exponent close parentheses equals$

For such questions, we should know u.v method.

$fraction numerator d over denominator d x end fraction open parentheses Tan to the power of negative 1 end exponent invisible function application square root of fraction numerator 1 minus c o s invisible function application x over denominator 1 plus c o s invisible function application x end fraction end root close parentheses equals$

We should know different formulas to solve such questions.

$fraction numerator d over denominator d x end fraction open parentheses Tan to the power of negative 1 end exponent invisible function application square root of fraction numerator 1 minus c o s invisible function application x over denominator 1 plus c o s invisible function application x end fraction end root close parentheses equals$

We should know different formulas to solve such questions.

$fraction numerator d over denominator d x end fraction open parentheses Tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator a minus x over denominator 1 plus infinity x end fraction close parentheses close parentheses equals$

For such questions, we should know different formulas.

$fraction numerator d over denominator d x end fraction open parentheses Tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator a minus x over denominator 1 plus infinity x end fraction close parentheses close parentheses equals$

For such questions, we should know different formulas.

$fraction numerator d over denominator d x end fraction open parentheses cosec to the power of negative 1 end exponent invisible function application open parentheses e to the power of 2 x plus 1 end exponent close parentheses close parentheses equals$

For such questions, we should know different formulas.

$fraction numerator d over denominator d x end fraction open parentheses cosec to the power of negative 1 end exponent invisible function application open parentheses e to the power of 2 x plus 1 end exponent close parentheses close parentheses equals$

For such questions, we should know different formulas.

$fraction numerator d over denominator d x end fraction open parentheses fraction numerator sin invisible function application x over denominator 1 plus c o s invisible function application x end fraction close parentheses equals$

The alternate method to solve this will be using u by method. It is method used in differentiation when we have a condition of numerator and denominator.

$fraction numerator d over denominator d x end fraction open parentheses fraction numerator sin invisible function application x over denominator 1 plus c o s invisible function application x end fraction close parentheses equals$

The alternate method to solve this will be using u by method. It is method used in differentiation when we have a condition of numerator and denominator.

Locus of the foot of perpendicular fr om (-2,3) to a variable line with $x$ intercept 2 is

Vertices of a triangle are (4, 3) , $left parenthesis 5 blank s i n blank capital theta comma blank 5 blank c o s blank capital theta right parenthesis comma$ $left parenthesis 5 blank c o s blank capital theta comma blank minus 5 blank s i n blank capital theta right parenthesis$ Then the locus of the orthocenter is

A line through A(-5, -4) meets the line x+3y+2=0, 2x+y+4=0 and x-y-5=0 at $B comma$ $C$ and $D$ respectively. If $left parenthesis fraction numerator 15 over denominator A B end fraction right parenthesis to the power of 2 end exponent plus left parenthesis fraction numerator 10 over denominator A C end fraction right parenthesis to the power of 2 end exponent equals left parenthesis fraction numerator 6 over denominator A D end fraction right parenthesis to the power of 2 end exponent$ then the equation of the line is

If $x subscript 1 end subscript comma x subscript 2 end subscript comma x subscript 3 end subscript$ as well as $y subscript 1 end subscript comma y subscript 2 end subscript comma y subscript 3 end subscript$ are in G.P with the same common ratio, then the points $left parenthesis x subscript 1 end subscript comma y subscript 1 end subscript right parenthesis comma$ $left parenthesis x subscript 2 end subscript comma y subscript 2 end subscript right parenthesis comma left parenthesis x subscript 3 end subscript comma y subscript 3 end subscript right parenthesis$ .