Math Examples of Derivatives

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).

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.

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.

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.

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.

Let's start with 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.

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

Let's start with 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. 

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). 

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.