The chain rule can be thought of as taking the derivative of the outer function (applied to the inner function) and multiplying it times the derivative of the inner function. The chain rule is often one of the hardest concepts for calculus students to understand. None of our rules will work on these functions and yet some of these functions are closer to the derivatives that we’re liable to run into than the functions in the first set. There are two forms of the chain rule. The following three problems require a more formal use of the chain rule. So even though the initial chain rule was fairly messy the final answer is significantly simpler because of the factoring. Instead we get $$1 - 5x$$ in both. $\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\,\,\frac{{du}}{{dx}}$, $$f\left( x \right) = \sin \left( {3{x^2} + x} \right)$$, $$f\left( t \right) = {\left( {2{t^3} + \cos \left( t \right)} \right)^{50}}$$, $$h\left( w \right) = {{\bf{e}}^{{w^4} - 3{w^2} + 9}}$$, $$g\left( x \right) = \,\ln \left( {{x^{ - 4}} + {x^4}} \right)$$, $$P\left( t \right) = {\cos ^4}\left( t \right) + \cos \left( {{t^4}} \right)$$, $$f\left( x \right) = {\left[ {g\left( x \right)} \right]^n}$$, $$f\left( x \right) = {{\bf{e}}^{g\left( x \right)}}$$, $$f\left( x \right) = \ln \left( {g\left( x \right)} \right)$$, $$T\left( x \right) = {\tan ^{ - 1}}\left( {2x} \right)\,\,\sqrt[3]{{1 - 3{x^2}}}$$, $$f\left( z \right) = \sin \left( {z{{\bf{e}}^z}} \right)$$, $$\displaystyle y = \frac{{{{\left( {{x^3} + 4} \right)}^5}}}{{{{\left( {1 - 2{x^2}} \right)}^3}}}$$, $$\displaystyle h\left( t \right) = {\left( {\frac{{2t + 3}}{{6 - {t^2}}}} \right)^3}$$, $$\displaystyle h\left( z \right) = \frac{2}{{{{\left( {4z + {{\bf{e}}^{ - 9z}}} \right)}^{10}}}}$$, $$f\left( y \right) = \sqrt {2y + {{\left( {3y + 4{y^2}} \right)}^3}}$$, $$y = \tan \left( {\sqrt[3]{{3{x^2}}} + \ln \left( {5{x^4}} \right)} \right)$$, $$g\left( t \right) = {\sin ^3}\left( {{{\bf{e}}^{1 - t}} + 3\sin \left( {6t} \right)} \right)$$. It’s now time to extend the chain rule out to more complicated situations. However, if you look back they have all been functions similar to the following kinds of functions. Now, differentiating the final version of this function is a (hopefully) fairly simple Chain Rule problem. However, the chain rule used to find the limit is different than the chain rule we use … As with the first example the second term of the inside function required the chain rule to differentiate it. What about functions like the following. a composite function). Use the chain rule to find $$\displaystyle \frac d {dx}\left(\sec x\right)$$. In the following discussion and solutions the derivative of a function h(x) will be denoted by or h'(x). Let’s look at an example of how these two derivative r but at the time we didn’t have the knowledge to do this. (x+1) but it will take longer, and also realise that when you use the product rule this time, the two functions are 'similiar'. Current time:0:00Total duration:2:27. chain rule composite functions composition exponential functions I want to talk about a special case of the chain rule where the function that we're differentiating has its outside function e to the x so in the next few problems we're going to have functions of this type which I call general exponential functions. The chain rule tells us how to find the derivative of a composite function. So, the power rule alone simply won’t work to get the derivative here. We use the chain rule when differentiating a 'function of a function', like f (g (x)) in general. MIT grad shows how to use the chain rule to find the derivative and WHEN to use it. Most of the examples in this section won’t involve the product or quotient rule to make the problems a little shorter. The square root is the last operation that we perform in the evaluation and this is also the outside function. Proving the chain rule. Get Better Unlike the previous problem the first step for derivative is to use the chain rule and then once we go to differentiate the inside function we’ll need to do the quotient rule. But sometimes it'll be more clear than not which one is preferable. But it's always ignored that even y=x^2 can be separated into a composition of 2 functions. So, the derivative of the exponential function (with the inside left alone) is just the original function. Eg: 45x^2/ (3x+4) Similarly, there are two functions here plus, there is a denominator so you must use the Quotient rule to differentiate. The chain rule is used to find the derivative of the composition of two functions. After factoring we were able to cancel some of the terms in the numerator against the denominator. • Solution 2. Here the outside function is the natural logarithm and the inside function is stuff on the inside of the logarithm. 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