Let’s try our hand at differentiating the simple example equation above. x2 + y2 - 5x + 8y + 2xy2 = 19 has two x terms: x2 and -5x. If we want to differentiate the equation, we’ll deal with these first, like this: x2 + y2 - 5x + 8y + 2xy2 = 19 (Bring the “2” exponent in x2 down as a coefficient, remove the x in -5x, and change the 19 to 0) 2x + y2 - 5 + 8y + 2xy2 = 0
In our running example, our equation now looks like this: 2x + y2 - 5 + 8y + 2xy2 = 0. We would perform this next y-differentiating step as follows: 2x + y2 - 5 + 8y + 2xy2 = 0 (Bring the “2” exponent in y2 down as a coefficient, remove the y in 8y, and place a “dy/dx” next to each). 2x + 2y(dy/dx) - 5 + 8(dy/dx) + 2xy2= 0
In our example, 2x + 2y(dy/dx) - 5 + 8(dy/dx) + 2xy2 = 0, we only have one term with both x and y — 2xy2. Since the x and y are multiplied by each other, we would use the product rule to differentiate as follows: 2xy2 = (2x)(y2)— set 2x = f and y2 = g in (f × g)’ = f’ × g + g’ × f (f × g)’ = (2x)’ × (y2) + (2x) × (y2)’ (f × g)’ = (2) × (y2) + (2x) × (2y(dy/dx)) (f × g)’ = 2y2 + 4xy(dy/dx) Adding this back into our main equation, we get 2x + 2y(dy/dx) - 5 + 8(dy/dx) + 2y2 + 4xy(dy/dx) = 0
In our example, 2x + 2y(dy/dx) - 5 + 8(dy/dx) + 2xy2 = 0, we only have one term with both x and y — 2xy2. Since the x and y are multiplied by each other, we would use the product rule to differentiate as follows: 2xy2 = (2x)(y2)— set 2x = f and y2 = g in (f × g)’ = f’ × g + g’ × f (f × g)’ = (2x)’ × (y2) + (2x) × (y2)’ (f × g)’ = (2) × (y2) + (2x) × (2y(dy/dx)) (f × g)’ = 2y2 + 4xy(dy/dx) Adding this back into our main equation, we get 2x + 2y(dy/dx) - 5 + 8(dy/dx) + 2y2 + 4xy(dy/dx) = 0
In our example, 2x + 2y(dy/dx) - 5 + 8(dy/dx) + 2xy2 = 0, we only have one term with both x and y — 2xy2. Since the x and y are multiplied by each other, we would use the product rule to differentiate as follows: 2xy2 = (2x)(y2)— set 2x = f and y2 = g in (f × g)’ = f’ × g + g’ × f (f × g)’ = (2x)’ × (y2) + (2x) × (y2)’ (f × g)’ = (2) × (y2) + (2x) × (2y(dy/dx)) (f × g)’ = 2y2 + 4xy(dy/dx) Adding this back into our main equation, we get 2x + 2y(dy/dx) - 5 + 8(dy/dx) + 2y2 + 4xy(dy/dx) = 0
In our example, we might simplify 2x + 2y(dy/dx) - 5 + 8(dy/dx) + 2y2 + 4xy(dy/dx) = 0 as follows: 2x + 2y(dy/dx) - 5 + 8(dy/dx) + 2y2 + 4xy(dy/dx) = 0 (2y + 8 + 4xy)(dy/dx) + 2x - 5 + 2y2 = 0 (2y + 8 + 4xy)(dy/dx) = -2y2 - 2x + 5 (dy/dx) = (-2y2 - 2x + 5)/(2y + 8 + 4xy) (dy/dx) = (-2y2 - 2x + 5)/(2(2xy + y + 4)
For example, let’s say that we want to find the slope at the point (3, -4) for our example equation above. To do this, we would substitute 3 for x and -4 for y, solving as follows: (dy/dx) = (-2y2 - 2x + 5)/(2(2xy + y + 4) (dy/dx) = (-2(-4)2 - 2(3) + 5)/(2(2(3)(-4) + (-4) + 4) (dy/dx) = (-2(16) - 6 + 5)/(2(2(3)(-4)) (dy/dx) = (-32) - 6 + 5)/(2(2(-12)) (dy/dx) = (-33)/(2(2(-12)) (dy/dx) = (-33)/(-48) = 3/48, or 0. 6875.
As a simple example, let’s say that we need to find the derivative of sin(3x2 + x) as part of a larger implicit differentiation problem for the equation sin(3x2 + x) + y3 = 0. If we think of sin(3x2 + x) as “f(x)” and 3x2 + x as “g(x)”, we can find the differentiation as follows: f’(g(x))g’(x) (sin(3x2 + x))’ × (3x2 + x)’ cos(3x2 + x) × (6x + 1) (6x + 1)cos(3x2 + x)
As a simple example, let’s say that we need to find the derivative of sin(3x2 + x) as part of a larger implicit differentiation problem for the equation sin(3x2 + x) + y3 = 0. If we think of sin(3x2 + x) as “f(x)” and 3x2 + x as “g(x)”, we can find the differentiation as follows: f’(g(x))g’(x) (sin(3x2 + x))’ × (3x2 + x)’ cos(3x2 + x) × (6x + 1) (6x + 1)cos(3x2 + x)
As a simple example, let’s say that we need to find the derivative of sin(3x2 + x) as part of a larger implicit differentiation problem for the equation sin(3x2 + x) + y3 = 0. If we think of sin(3x2 + x) as “f(x)” and 3x2 + x as “g(x)”, we can find the differentiation as follows: f’(g(x))g’(x) (sin(3x2 + x))’ × (3x2 + x)’ cos(3x2 + x) × (6x + 1) (6x + 1)cos(3x2 + x)
For example, let’s say that we’re trying to differentiate x3z2 - 5xy5z = x2 + y3. First, let’s differentiate with respect to x and insert (dz/dx). Don’t forget to apply the product rule where appropriate! x3z2 - 5xy5z = x2 + y3 3x2z2 + 2x3z(dz/dx) - 5y5z - 5xy5(dz/dx) = 2x 3x2z2 + (2x3z - 5xy5)(dz/dx) - 5y5z = 2x (2x3z - 5xy5)(dz/dx) = 2x - 3x2z2 + 5y5z (dz/dx) = (2x - 3x2z2 + 5y5z)/(2x3z - 5xy5) Now, let’s do the same for (dz/dy) x3z2 - 5xy5z = x2 + y3 2x3z(dz/dy) - 25xy4z - 5xy5(dz/dy) = 3y2 (2x3z - 5xy5)(dz/dy) = 3y2 + 25xy4z (dz/dy) = (3y2 + 25xy4z)/(2x3z - 5xy5)
