05 The equation of a curve is given by y^3+ xy = 2. ( Find an expression for dy/dx in terms of x and y. Hence find the gradient of the curve at (1, 1) and the equation of the tangent to the curve at that point.
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Answer:
Gradient of a curve
y
=
x
3
+
7
is given by its derivative
d
y
d
x
.
As
y
=
x
3
+
7
,
d
y
d
x
=
3
x
2
x
=
1
is
d
y
d
x
=
3
⋅
1
2
=
3
⋅
1
=
3
(
x
=
1
)
=
3
Explanation:
Given -
y
=
x
3
+
7
.
d
y
d
x
=
3
x
2
At
x
=
1
At
x
=
1
;
d
y
d
x
=
3
(
1
2
)
=
3
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