The distance of the line 2x – 3y = 4 from the point (1, 1) in the direction of the line x + y = 1 is

The image of the point A (1, 2) by the line mirror y = x is the point B and the image of B by line mirror y = 0 is the point

If $y equals T a n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator square root of 1 plus a to the power of 2 end exponent x to the power of 2 end exponent end root minus 1 over denominator a x end fraction close parentheses text then end text open parentheses 1 plus a to the power of 2 end exponent x to the power of 2 end exponent close parentheses y to the power of parallel to end exponent plus 2 a to the power of 2 end exponent x y to the power of ´ end exponent equals$

For such questions, we should know different trigonometric formulas. We should simplify the function first before finding derivatives.

If $y equals T a n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator square root of 1 plus a to the power of 2 end exponent x to the power of 2 end exponent end root minus 1 over denominator a x end fraction close parentheses text then end text open parentheses 1 plus a to the power of 2 end exponent x to the power of 2 end exponent close parentheses y to the power of parallel to end exponent plus 2 a to the power of 2 end exponent x y to the power of ´ end exponent equals$

For such questions, we should know different trigonometric formulas. We should simplify the function first before finding derivatives.

If $y equals c o s to the power of negative 1 end exponent invisible function application open parentheses fraction numerator a to the power of 2 end exponent minus x to the power of 2 end exponent over denominator a to the power of 2 end exponent plus x to the power of 2 end exponent end fraction close parentheses plus s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 2 a x over denominator a to the power of 2 end exponent plus x to the power of 2 end exponent end fraction close parentheses text then end text fraction numerator d y over denominator d x end fraction equals$

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

If $y equals c o s to the power of negative 1 end exponent invisible function application open parentheses fraction numerator a to the power of 2 end exponent minus x to the power of 2 end exponent over denominator a to the power of 2 end exponent plus x to the power of 2 end exponent end fraction close parentheses plus s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 2 a x over denominator a to the power of 2 end exponent plus x to the power of 2 end exponent end fraction close parentheses text then end text fraction numerator d y over denominator d x end fraction equals$

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