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Combining Derivative Rules

In calculus, especially when working with derivatives, it’s important to understand three main rules: the Product Rule, Quotient Rule, and Chain Rule. These rules help us find the derivatives of complicated functions. They allow us to break down expressions that involve different mathematical operations in an easy way.

Basic Rules to Remember

  1. Product Rule: The Product Rule helps when you’re dealing with two functions multiplied together, like u(x)u(x) and v(x)v(x). The rule says: ddx[u(x)v(x)]=u(x)v(x)+u(x)v(x).\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x). This means you first find the derivative of the first function and multiply it by the second function, then add it to the first function multiplied by the derivative of the second function.

  2. Quotient Rule: When you have a fraction, you’ll use the Quotient Rule. If u(x)u(x) and v(x)v(x) are functions, the rule goes like this: ddx[u(x)v(x)]=u(x)v(x)u(x)v(x)[v(x)]2.\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}. This helps us differentiate when one function is divided by another.

  3. Chain Rule: The Chain Rule is really useful for composite functions, which are functions inside other functions. If you have y=f(g(x))y = f(g(x)), the derivative is: dydx=f(g(x))g(x).\frac{dy}{dx} = f'(g(x)) \cdot g'(x). This means you find the derivative of the outer function and multiply it by the derivative of the inner function.

How to Use These Rules

When figuring out which rule to use, here are some steps to help:

  • Look at the Function's Structure: Check if the function is a product, quotient, or composite of functions.

  • Break It Down: Simplify the function into smaller parts. This might mean rewriting it to make it easier to apply the rules.

  • Use Rules in Order: Sometimes you need to use multiple rules one after another. Figure out which rule to use first and then proceed with the others as needed.

Examples of Using the Rules

Here’s a straightforward example: y=(3x2+2)(sin(x)).y = (3x^2 + 2)(\sin(x)). Using the Product Rule, we get: y=(6x)(sin(x))+(3x2+2)(cos(x)).y' = (6x)(\sin(x)) + (3x^2 + 2)(\cos(x)).

Now for a more complicated example: y=(x2+1)(ln(x))ex.y = \frac{(x^2 + 1)(\ln(x))}{e^x}. Here, we apply the Quotient Rule: y=(2xln(x)+x2+1x)(ex)((x2+1)(ln(x)))(ex)(ex)2.y' = \frac{(2x \ln(x) + \frac{x^2 + 1}{x})(e^x) - ((x^2 + 1)(\ln(x)))(e^x)}{(e^x)^2}.

Try These Practice Problems!

To help you practice these rules, here are two problems:

  1. Differentiate the function: y=(x3+3)(tan(x)).y = (x^3 + 3)(\tan(x)).

  2. Find the derivative of: y=sin(x2)(x3)x+1.y = \frac{\sin(x^2)(x^3)}{x + 1}.

By understanding these rules and practicing, you will get better at finding derivatives in calculus. Keep practicing, and soon you’ll confidently tackle even the most complicated problems!

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Combining Derivative Rules

In calculus, especially when working with derivatives, it’s important to understand three main rules: the Product Rule, Quotient Rule, and Chain Rule. These rules help us find the derivatives of complicated functions. They allow us to break down expressions that involve different mathematical operations in an easy way.

Basic Rules to Remember

  1. Product Rule: The Product Rule helps when you’re dealing with two functions multiplied together, like u(x)u(x) and v(x)v(x). The rule says: ddx[u(x)v(x)]=u(x)v(x)+u(x)v(x).\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x). This means you first find the derivative of the first function and multiply it by the second function, then add it to the first function multiplied by the derivative of the second function.

  2. Quotient Rule: When you have a fraction, you’ll use the Quotient Rule. If u(x)u(x) and v(x)v(x) are functions, the rule goes like this: ddx[u(x)v(x)]=u(x)v(x)u(x)v(x)[v(x)]2.\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}. This helps us differentiate when one function is divided by another.

  3. Chain Rule: The Chain Rule is really useful for composite functions, which are functions inside other functions. If you have y=f(g(x))y = f(g(x)), the derivative is: dydx=f(g(x))g(x).\frac{dy}{dx} = f'(g(x)) \cdot g'(x). This means you find the derivative of the outer function and multiply it by the derivative of the inner function.

How to Use These Rules

When figuring out which rule to use, here are some steps to help:

  • Look at the Function's Structure: Check if the function is a product, quotient, or composite of functions.

  • Break It Down: Simplify the function into smaller parts. This might mean rewriting it to make it easier to apply the rules.

  • Use Rules in Order: Sometimes you need to use multiple rules one after another. Figure out which rule to use first and then proceed with the others as needed.

Examples of Using the Rules

Here’s a straightforward example: y=(3x2+2)(sin(x)).y = (3x^2 + 2)(\sin(x)). Using the Product Rule, we get: y=(6x)(sin(x))+(3x2+2)(cos(x)).y' = (6x)(\sin(x)) + (3x^2 + 2)(\cos(x)).

Now for a more complicated example: y=(x2+1)(ln(x))ex.y = \frac{(x^2 + 1)(\ln(x))}{e^x}. Here, we apply the Quotient Rule: y=(2xln(x)+x2+1x)(ex)((x2+1)(ln(x)))(ex)(ex)2.y' = \frac{(2x \ln(x) + \frac{x^2 + 1}{x})(e^x) - ((x^2 + 1)(\ln(x)))(e^x)}{(e^x)^2}.

Try These Practice Problems!

To help you practice these rules, here are two problems:

  1. Differentiate the function: y=(x3+3)(tan(x)).y = (x^3 + 3)(\tan(x)).

  2. Find the derivative of: y=sin(x2)(x3)x+1.y = \frac{\sin(x^2)(x^3)}{x + 1}.

By understanding these rules and practicing, you will get better at finding derivatives in calculus. Keep practicing, and soon you’ll confidently tackle even the most complicated problems!

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