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# Math Examples of Derivatives

• ## Find out the expression for dy / dx, if the given equation is y4 + 2x2y2 + 6x2 = 7?

On differentiating the equation with respect to x.
Let's understand the differentiation of each term separately.
First term
d/dxy4 = 4y3 (dy/dx),
Second term
d/dx2x2y2 = (2x2) 2y(dy/dx) + (y2) (4x),
Third term
d/dx2x2y2 = 4x2 y(dy/dx) + 4xy2 ,
[Since the formula for differentiation of u.v is u (dv/dx) + v (du /dx),
Fourth term
d/dx (6x2) = 12x,
and the constant
d/dx(7) = 0
Thus the differentiation of the given equation is expressed as
d/dxy4 + 2x2y2 + 6x2) = d/dx (7),
[4y3 (dy/dx)] + [(2x2) 2y(dy/dx) + (y2) (4x)] + [12x] = 0,
On simplifying the equation
(4y3 + 4x2y) dy / dx = - 4xy2 – 12x,
Thus the required solution is
dy / dx = (-4xy2 – 12x) / (4y3 + 4x2y),
dy / dx = (-xy2 – 3x) / (y3 + x2y).

## Determine the derivative of the function which is implicitly is defined by the relation sin (2x – 7y) = 16y?

We have to calculate the derivative dy/dx for the function of ‘x’ that implicitly is defined by sin (2x – 7y) = 16y.
d/dx[sin (2x – 7y)] = d/dx (16y),
cos (2x – 7y) d/dx (2x – 7y) = 16 (dy/dx),
cos (2x – 7y) (2 – 7 (dy/dx)) = 16 dy/dx,
2 cos (2x – 7y) – 7 cos (2x – 7y) dy/dx = 16 dy/dx,
-7 cos (2x – 7y) dy/dx – 16 dy/dx = -2 cos (2x – 7y),
Thus the value of the derivative is,
dy/dx = -2 cos (2x – 7y) / -7 cos (2x – 7y) – 16.

## Determine the derivative of the function that implicitly is defined by the relation given as √(5y – 3x) = x – 6y?

The derivative of the given equation is
d/dx [√(5y – 3x)] = d/dx [x – 6y],
( ½ ) (5y – 3x)(-1/2) d/dx (5y – 3x) = 1 – 6 (dy/dx),
( ½ ) (5y – 3x)(-1/2) (5 dy/dx – 3) = 1 – 6 (dy/dx),
5.( ½ ) (5y – 3x)(-1/2) dy/dx – 3.( ½ ) (5y – 3x)(-1/2) = 1 – 6 (dy/dx),
(5/2) (5y – 3x)(-1/2) dy/dx + 6 (dy/dx) = 1 + (3/2) (5y – 3x)(-1/2),
Thus the derivative of the equation is,
dy/dx = 1 + (3/2) (5y – 3x)(-1/2) / (5/2) (5y – 3x)(-1/2) + 6.

## Determine the slope of the tangent at the point (2,-1) for the given curve 2y + 5 – x2 – y3 = 0?

On differentiating the equation of the curve with respect to ‘x’. Here we are considering that ‘y’ is a function of ‘x’.
2 (dy/dx) + 0 + 2x – 3y2 (dy/dx) = 0,
On collecting like terms we get,
(2 – 3y2) dy/dx = 2x,
So
dy/dx = 2x / 2 - 3y2,
Thus at the Point x= 2 and y = -1, The derivative
dy/dx = 2 * 2 / 2 – 3(-1)2 = -4,
Thus the Slope of the Tangent at the point (2,-1) is -4.
Thus the equation of the tangent line is-
y – y1 = m (x – x1),
at point (2,3).
y – (-1) = (-4) (x – 2),
y + 1 = -4x + 2,
4x + y = 1.

## Calculate the equation of the tangent line to the ellipse 25x2 + y2 = 109, at the given point (2,3)?

In the given problem let's consider y as a function of x that is
25x2 + y(x)2 = 109,
On differentiating this equation we get
d/dx [25x2 + y(x)2] = 0,
50 x + 2y (dy / dx) = 0,
Thus we get first derivative of y as
dy / dx = -(25x) /y,
As we need to calculate the Tangent line at the Point (2,3) thus
dy / dx = -(25 * 2) / 3,
dy / dx = -50 / 3,
Thus the equation of the tangent line
y – y1 = m (x – x1),
at point (2,3),
y – 3 = (-50 / 3) (x – 2),
or on simplifying the equation,
50x + 3y = 109,
This is the equation of the tangent line we want.

## Suppose that ‘y’ is a function of ‘x’ then calculate y' = dy / dx for cos2x + cos2y = cos (2x + 2y)?

On differentiating both sides of the equations we get:
D (cos2x + cos2y) = D (cos (2x + 2y)),
D (cos2x ) + D (cos2y) = D (cos (2x + 2y)),
2 cos x D(cos x) + (2 cos y) D (cos y) = - sin (2x + 2y) D (2x + 2y),
On differentiating
2 cos x (- sinx) + 2 cos y (-sin y) y' = - sin (2x + 2y) D(2x + 2y),
Now for solving the value of y' we get
-2 cos x sin x – 2y' cos y sin y = - 2sin (2x + 2y) – 2y' sin (2x + 2y),
2y' sin (2x + 2y) – 2y' cos y sin y = - 2sin (2x + 2y) + 2 cos x sin x,
Now factor out y' to get the value of y'
y'[2sin (2x + 2y) – 2 cos y sin y] = 2 cos x sin x - 2sin (2x + 2y),
y' = 2 cos x sin x - 2sin (2x + 2y) / 2sin (2x + 2y) – 2 cos y sin y,
y' = cos x sin x - sin (2x + 2y) / sin (2x + 2y) – cos y sin y.

## Suppose that ‘y’ is a function of ‘x’ then find out y' = dy / dx for exy = e4x – e5y?

Beginning with exy = e4x – e5y.
On differentiating both sides of the equation, we get
D(exy) = D(e4x – e5y),
D(exy) = D(e4x) – D(e5y),
(exy) D(xy) = (e4x) D(4x) – (e5y) D(5y),
(exy) (xy' + (1)y) = (e4x) (4) – (e5y) (5y'),
Now to solve for y'
x (exy) y' + y (exy) = 4 (e4x) – 5 (e5y) y',
x (exy) y' + 5 (e5y) y' = 4 (e4x) - y (exy),
Now factor out y' to get the value of y'
y' [x (exy) + 5 (e5y)] = 4 (e4x) - y (exy),
or
y' = 4 (e4x) - y (exy) / [x (exy) + 5 (e5y)].

## Suppose that ‘y’ is a function of ‘x’ then calculate y' = dy / dx for y = x2y3 + x3y2?

On differentiating both sides of the equation we get:
D(y) = D(x2y3 + x3y2),
D(y) = D(x2y3) + D(x3y2),
On using the Product rule twice
y' = x2 D(y3) + D(x2y3) + x3 D(y2) + D(x3y2),
On using the Chain Rule on d(y3) and D(y2)
y' = x2 (3y2y') + (2x)y3 + x3(2yy') + (3x2)y2,
y' = 3x2y2y' + 2xy3 + 2x3yy' + 3x2y2,
To solve now for the value of y'
y' – 3x2y2y' – 2x3yy' = 2xy3 + 3x2y2,
Now factor out y'
y' [1 - 3x2y2 – 2x3y] = 2xy3 + 3x2y2,
or
y' = 2xy3 + 3x2y2 / 1 - 3x2y2 – 2x3y.

## Suppose that ‘y’ is a function of ‘x’ then calculate the value of y' = dy / dx for y = sin (3x + 4y)?

Let's start with the given equation y = sin (3x + 4y).
On differentiating both sides of the equation, we get:
D(y) = D sin (3x + 4y)
As we know that implicit solutions are the special case of Chain Rule, thus on solving Dsin (3x + 4y) with the help of chain rule
y' = cos (3x + 4y) D(3x + 4y)
y' = cos (3x + 4y) (3 + 4y')
Thus to solve for y'
y' = 3 cos (3x + 4y) + 4y' cos (3x + 4y),
y' – 4y' cos (3x + 4y) = 3 cos (3x + 4y),
Factor out the value of y'
y' [1 – 4 cos (3x + 4y)] = 3 cos (3x + 4y),
or
y' = 3 cos (3x + 4y) / 1 – 4 cos (3x + 4y).

## Suppose that ‘y’ is a function of ‘x’ then calculate y' = dy / dx for (x – y)2 = x + y -1?

Let's have an starting with (x – y)2 = x + y -1.
On differentiating both sides of the equation we get
D(x – y)2 = D(x + y -1),
D(x – y)2 = D(x) + D(y) – D(1),
As we know that implicit solutions are the special case of Chain Rule, thus on solving this with the help of chain rule
2(x – y) D(x – y) = 1 + y' – 0,
2( x – y) (1 – y') = 1 + y',
To solve the equation for y', we get
2(x – y) – 2(x – y) y' = 1 + y',
-2(x – y) y' – y' = 1 – 2(x – y),
Now factor out y'
y'[-2(x – y) – 1] = 1 – 2(x – y),
and
y' = 1 – 2(x – y) / -2(x – y) – 1,
y' = (2y – 2x + 1) / (2y – 2x – 1),
Thus this is the calculated y' for the given problem.
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