Application Of Derivatives

Question

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Let y be an implicit function of x defined by x^2x – 2x^xcoty – 1 = 0. Then y′ (1) equals

-1

1

log 2

-log 2

Solution

-1

When x = 1, y=π/2
=(x^x– cot y)²= cosec²y
x^x = cot y + |cosec y|
when x = 1, y=π/2

⇒ x^x = cot y + cosec y
diff. w.r.t. to x
$straight x to the power of straight x space left parenthesis 1 space plus space lnx right parenthesis space equals space left parenthesis negative cosec squared straight y space minus cosecy space cot space straight y right parenthesis space dy over dx When space straight x space equals 1 space and space straight y space equals straight pi over 2 dy over dx space equals space minus space 1$

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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Application Of Derivatives

Let y be an implicit function of x defined by x^2x – 2x^xcoty – 1 = 0. Then y′ (1) equals

-1

1

log 2

-log 2

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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NCERT Sample Papers

Entrance Exams Preparation

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For Daily Free Study Material Join wiredfaculty

Application Of Derivatives

Let y be an implicit function of x defined by x2x – 2xxcoty – 1 = 0. Then y′ (1) equals -1 1 log 2 -log 2

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Let y be an implicit function of x defined by x^2x – 2x^xcoty – 1 = 0. Then y′ (1) equals

-1

1

log 2

-log 2