Question

find the equation of tangent to the curve y=x³ at (1,1)​

Answer 1

Answer:

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Step-by-step explanation:

Answer

On differentiating with respect to x, we get:

dx

dy

=3x

2

(

dx

dy

)

(1,1)

=3(1)

2

=3

Thus, the slope of the tangent at (1,1) is 3 and the equation of the tangent is given as,

y−1=3(x−1)⇒y=3x−2

The slope of the normal at (1,1) is

Slope of the tangent at(1,1)

−1

=

3

−1

.

Therefore, the equation of the normal at (1,1) is given as,

y−1=−

3

1

(x−1)⇒x+3y−4=0

Answer 2

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