DIFFERENTIAL CALCULUS | THE CHAIN RULE

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The chain rule is a fundamental theorem in calculus that allows us to find the derivative of a composite function. A composite function is a function that is formed by taking one function and plugging it into another function. The chain rule tells us how to find the derivative of such a function.

In calculus, the derivative of a function gives us the rate of change of that function at a particular point. The chain rule tells us how the rate of change of a composite function depends on the rate of change of its individual parts.

The chain rule can be stated as follows:

If y = f(g(x)) is a composite function, where g(x) is a differentiable function and f(u) is a differentiable function of u, then the derivative of y with respect to x is given by:

$\frac{dy}{dx}=\frac{df}{du}.\frac{du}{dx}$

where du/dx is the derivative of g(x) with respect to x and df/du is the derivative of f(u) with respect to u, evaluated at u = g(x).

In other words, the chain rule tells us that to find the derivative of a composite function, we first find the derivative of the outer function (f) with respect to its input (u), and then multiply that by the derivative of the inner function (g) with respect to its input (x).

USES OF CHAIN RULE

The chain rule is an important tool in calculus, as it allows us to find the derivative of more complex functions by breaking them down into simpler parts. It is used in a wide range of applications, including physics, engineering, and economics.

Uses in Physics:

Kinematics: The chain rule is used to find the derivative of position, velocity, and acceleration with respect to time in one, two, or three dimensions. For example, if an object is moving in two dimensions with position (x,y) given by a function of time, then the velocity components (dx/dt, dy/dt) can be found using the chain rule.
Thermodynamics: The chain rule is used to find the rate of change of temperature, entropy, or other thermodynamic variables with respect to time. For example, the rate of change of temperature with respect to time can be expressed as the product of the specific heat capacity and the rate of change of internal energy with respect to temperature.
Electromagnetism: The chain rule is used to find the gradient, divergence, and curl of vector fields. For example, the divergence of an electric field can be found by taking the dot product of the gradient operator with the electric field, using the chain rule to differentiate the components of the electric field.
Quantum mechanics: The chain rule is used to find the derivative of wave functions with respect to time or position. For example, in the Schrödinger equation, the time derivative of the wave function is proportional to the second derivative of the wave function with respect to position, using the chain rule to differentiate the complex-valued wave function.

Uses in Engineering:

Control Systems: The chain rule is used to find the derivative of signals in control systems. For example, in a feedback control system, the derivative of the error signal with respect to time is used to adjust the control input to the system. The chain rule can be used to find the derivative of the error signal with respect to the system inputs.
Electrical Engineering: The chain rule is used to find the derivative of signals in electrical engineering. For example, the chain rule is used to find the derivative of voltage or current signals in electrical circuits. The chain rule is also used to find the derivative of the magnetic and electric fields in electromagnetic systems.
Mechanical Engineering: The chain rule is used to find the derivative of velocities and accelerations in mechanical engineering. For example, the chain rule is used to find the derivative of the angular velocity and angular acceleration of rotating machinery.
Robotics: The chain rule is used to find the derivative of the motion of robotic manipulators. For example, the chain rule is used to find the derivative of the position and orientation of a robot's end-effect or with respect to its joint angles.

Uses in Economics:

Marginal Analysis: The chain rule is used to find the marginal rate of substitution, marginal revenue, and marginal cost in economics. For example, the marginal rate of substitution measures the rate at which a consumer is willing to substitute one good for another. The chain rule can be used to find the derivative of the utility function with respect to the quantities of the two goods.
Production Functions: The chain rule is used to find the elasticity of production with respect to inputs. For example, the elasticity of production measures the percentage change in output due to a 1% change in the inputs. The chain rule can be used to find the derivative of the production function with respect to the inputs.
Econometrics: The chain rule is used to find the derivatives of econometric models with respect to their parameters. For example, in linear regression, the chain rule can be used to find the partial derivatives of the sum of squared errors with respect to the coefficients of the regression equation.
Finance: The chain rule is used to find the derivatives of financial models with respect to their inputs. For example, the Black-Scholes model for pricing options involves the partial derivatives of the option price with respect to the underlying asset price and volatility, using the chain rule to differentiate the Black-Scholes equation.

Sample Step Guide on How to Do it:

1. Given the function below:

$y=(2x^^{2}+3x-1)^{3}$

We want to find the derivative of y with respect to x.

Using the algebraic chain rule, we can rewrite this function as a composition of two functions:

$y=f(u)^{3},\ where\ f(u)=2u^{2}+3u-1,\ and\ u=x^2$

To find dy/dx, we use the chain rule formula:

$\frac{dy}{dx}=\frac{df}{du}.\frac{du}{dx}$

First, we find the derivative of f(u) with respect to u:

$\frac{df}{du}=4u+3,{\color{DarkRed} ({\color{Red} i.e:2*2u^{2-1}+1*3u^{1-1}+0})}$

Next, we find the derivative of u with respect to x:

$\frac{du}{dx}=2x$

Substituting these values into the chain rule formula, we get:

$\frac{dy}{dx}=(4u + 3)(2x)$

Now we need to express u in terms of x. Recall that $u=x^2$ . Substituting this value into the equation above, we get:

$\frac{dy}{dx}={(4x^2+3)}(2x)$

Simplifying this expression, we get:

$\frac{dy}{dx}={(8x^3+6x)}$

Therefore, the derivative of y with respect to x is:

$\frac{dy}{dx}=8x^3+6x$ , final answer

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2. Given the function below:

$y=\sqrt{(3x^2 + 1)^5}$

We want to find the derivative of y with respect to x.

We can rewrite the function as a composition of two functions:

$y=f(u)^{\frac{1}{2}},\ where \ f(u) = (3u + 1)^5, and\ u = x^2$

To find dy/dx, we use the chain rule formula:

$\frac{dy}{dx}=\frac{\frac{df}{du}.\frac{du}{dx}}{2\sqrt{f(u)}}$

First, we find the derivative of f(u) with respect to u:

$\frac{df}{du}=15(3u+1)^4$

Next, we find the derivative of u with respect to x:

$\frac{du}{dx}=2x$

Substituting these values into the chain rule formula, we get:

$\frac{dy}{dx}=\frac{15(3u + 1)^4.(2x)}{2\sqrt{f(u)}}$

Now we need to express u in terms of x. Recall that $u = x^2$ . Substituting this value into the equation above, we get:

$\frac{dy}{dx}=\frac{15(3x^2 + 1)^4.({\color{Red} 2}x)}{{\color{Red} 2}\sqrt{3x^2 + 1)^5}}$

Simplifying this expression, we get:

$\frac{dy}{dx}=\frac{15x(3x^2 + 1)^4}{{\color{Red} }\sqrt{3x^2 + 1)^5}}$

Therefore, the derivative of y with respect to x is:

$\frac{dy}{dx}=\frac{15x(3x^2 + 1)^4}{{\color{Red} }\sqrt{3x^2 + 1)^5}}$ , final answer

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3. Given that

$y= (3x^3 - 2x^2 + 5x - 1)^2$ , Find its derivative.

Again, we want to find the derivative of y with respect to x.

We can rewrite the function as a composition of two functions:

$y = f(u)^2,\ where \ f(u) = 3u^3 - 2u^2 + 5u - 1, and \ u = x^3.$

To find dy/dx, we use the chain rule formula:

$\frac{dy}{dx}=\frac{df}{du}.\frac{du}{dx}.2f(u)$

First, we find the derivative of f(u) with respect to u:

$\frac{df}{du} = 9u^2 - 4u + 5$

Next, we find the derivative of u with respect to x:

$\frac{du }{dx}= 3x^2$

Substituting these values into the chain rule formula, we get:

$\frac{dy}{dx} = (9u^2 - 4u + 5)(3x^2) * 2(3u^3 - 2u^2 + 5u - 1)^2$

Now we need to express u in terms of x. Recall that $u = x^3$ . Substituting this value into the equation above, we get:

$\frac{dy}{dx} = (9x^6 - 4x^3 + 5)(3x^2) * 2(3x^9 - 2x^6 + 5x^3 - 1)^2$

Simplifying this expression, we get:

$\frac{dy}{dx} = 6x^5(27x^{18} - 36x^{12} + 75x^9 - 10x^6 + 25x^3 - 2)$

Therefore, the derivative of y with respect to x is:

$\frac{dy}{dx} = 6x^5(27x^{18} - 36x^{12} + 75x^9 - 10x^6 + 25x^3 - 2)$ , final answer

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4. Given the function below,

$y = (2x^2 + 1)^{3x - 1}$

We want to find the derivative of y with respect to x.

At first glance, this function might not look like a composite function, but we can still use the chain rule by rewriting it as:

$y = f(u)^v, \ where \ f(u) = 2u + 1, v = 3x - 1, and \ u = x^2.$

To find dy/dx, we use the chain rule formula:

$\frac{dy}{dx}=\frac{df}{du}.\frac{du}{dx}*v*f(u)^{v-1}$

First, we find the derivative of f(u) with respect to u:

$\frac{df}{du} = 2$

Next, we find the derivative of u with respect to x:

$\frac{du}{dx} = 2x$

Finally, we need to find v and f(u) in terms of x. Recall that v = 3x - 1 and $u = x^2$ . Substituting these values into the equation above, we get:

$\frac{dy}{dx} = (2)(2x) * (3x - 1) * (2x^2 + 1)^{3x - 2}$

Simplifying this expression, we get:

$\frac{dy}{dx} = 4x(3x - 1)(2x^2 + 1)^{3x - 2}$

Therefore, the derivative of y with respect to x is:

$\frac{dy}{dx} = 4x(3x - 1)(2x^2 + 1)^{3x - 2}$ , final answer

This shows how the algebraic chain rule can be used to find the derivative of a composite function. We can break down the function into simpler pieces and use the chain rule formula to find the derivative of the overall function.

Sample Problems with Solution:

1. Find the derivative of

$y = (2x - 1)^3$ .

Solution:

Let u = 2x - 1. Then $y = u^3$ . Using the chain rule formula, we have:

$\frac{dy}{dx} = (\frac{df}{du})(\frac{du}{dx}) = (3u^2)(2) = 6(2x - 1)^2.$

Therefore, the derivative of y with respect to x is,

$\frac{dy}{dx} = 6(2x - 1)^2.$

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2. Find the derivative of

$y=\sqrt{5x - 1}$

Solution:

Let u = 5x - 1. Then $y=\sqrt{u}$ . Using the chain rule formula, we have:

$\frac{dy}{dx} = (\frac{df}{du})(\frac{du}{dx}) = (1/2\sqrt{u})(5) = 5/(2\sqrt{5x - 1})=\frac{5}{2\sqrt{5x - 1}$

Therefore, the derivative of y with respect to x is

$\frac{dy}{dx} = \frac{5}{2\sqrt{5x - 1}$

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3. Find the derivative of

$y = (2x^2 - 1)^4.$

Solution:

Let $u = 2x^2 - 1$ . Then $y = u^4$ . Using the chain rule formula, we have:

$\frac{dy}{dx} = (\frac{df}{du})(\frac{du}{dx}) = (4u^3)(4x) = 16x(2x^2 - 1)^3.$

Therefore, the derivative of y with respect to x is

$\frac{dy}{dx} = (2x^2 - 1)^4= 16x(2x^2 - 1)^3.$

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4. Find the derivative of

$y = e^{3x - 1}$

Solution:

Let $u = 3x - 1$ . Then $y = e^u$ . Using the chain rule formula, we have:

$\frac{dy}{dx} = (\frac{df}{du})(\frac{du}{dx}) = e^u(3) = 3e^{3x - 1}$

Therefore, the derivative of y with respect to x is

$\frac{dy}{dx} = e^{3x - 1} = 3e^{3x - 1}$

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5 .Find the derivative of

$y = ln(4x^2 - 3)$

Solution:

Let $u = 4x^2 - 3$ . Then

$y = ln(u)$

Using the chain rule formula, we have:

$\frac{dy}{dx} = (\frac{df}{du})(\frac{du}{dx}) = (\frac{1}{u})(8x) = \frac{8x}{4x^2 - 3}$

Therefore, the derivative of y with respect to x is

$\frac{dy}{dx} = ln(4x^2 - 3)= \frac{8x}{4x^2 - 3}$

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Problems with Partial Solution:

1. Find the derivative of

$y = (2x + 1)^5.$

Solution:

Let u = 2x + 1.

Then

$y = u^5$ .

To derive, use the chain rule formula provided below;

$\frac{dy}{dx} = (\frac{df}{du})(\frac{du}{dx})$

(Please continue to finish..)

2. Find the derivative of

$f(x) = (x^2 + 1)^4$

Solution:

Let $u = x^2 + 1$ .

Then

$f(x) = u^4$ .

Applying chain rule, we get

$f'(x) = 4u^3 * 2x = {\color{Red} ??}$

(Please continue to finish..)

3. Find the derivative of

$f(x) = sin(x^2 + 1)$

Solution:

$Let \ u = x^2 + 1.$

Then,

$f(x) = sin(u)$

Applying chain rule, we get

$f'(x) = cos(u) * 2x = {\color{Red} ??}$

(Please continue to finish..)

4. Find the derivative of

$f(x) = e^{(5x^3 - 2x)}$

Solution:

Let,

$u = 5x^3 - 2x$

Then

$f(x) = e^u$

Applying chain rule, we get

$f'(x) = e^u * (15x^2 - 2)$

(Please continue to finish..)

5. Find the derivative of

$f(x) = (x^2 + 1)^2 * (2x - 3)^3$

Solution:

Let

$u=x^2+1$

and

$v = 2x - 3$

Then

$f(x) = u^2 * v^3$

Applying chain rule, we get

$f'(x) = 2u * du/dx * v^3 + u^2 * 3v^2 * dv/dx = 2(x^2 + 1) * 2x * (2x - 3)^3 + (x^2 + 1)^2 * 3(2x - 3)^2 * 2 ={\color{Red} ??}$

(Please continue to finish..)

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Simple Quiz

Write your answer in the comments section.

Find the derivative of f(x) = e^(3x^2 - 5x + 1).
Find the derivative of f(x) = ln(sin(x)).
Find the derivative of f(x) = (x^3 + 2x - 1)^5.
Find the derivative of f(x) = sin(2x + cos(x)).
Find the derivative of f(x) = ln(x^2 + 3x - 2).
Find the derivative of f(x) = e^(x^2 - 3x) * cos(4x).
Find the derivative of f(x) = (2x + 1)^2 * (3x - 4)^3.
Find the derivative of f(x) = tan(x^2 - 1).
Find the derivative of f(x) = (3x^2 + 4x + 2) / (2x^2 - 5x + 1).
Find the derivative of f(x) = ln(cos(x) + 1).

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