Calculus:
Derivative: Chain Rule

1. 
\(\frac{d}{dx} {(\cos{({x}^{3})})}^{2}\)
 Solution
2. 
\(\frac{d}{dx} \ln{({x}^{2}+1)}\)
 Solution
3. 
\(\frac{d}{dx} \sqrt{13{x}^{2}-5x+8}\)
 Solution
4. 
\(\frac{d}{dx} \cos{(\ln{({x}^{3})})}\)
 Solution
5. 
\(\frac{d}{dx} x+\sqrt{\csc{x}}\)
 Solution
6. 
\(\frac{d}{dx} {x}^{3}+\sec{({x}^{4})}\)
 Solution
7. 
\(\frac{d}{dx} \sin^{3x}x\)
 Solution
8. 
\(\frac{d}{dy} \tan{({e}^{-3y})}\)
 Solution

The Chain Rule for Derivatives: Introduction

In calculus, students are often asked to find the “derivative” of a function. You can think of this graphically: the derivative of a function is the slope of the tangent line to the function at the given point. Still seems confusing? Let's take an example: You should know that the slope of any horizontal line is 0. Therefore, the slope of y=c where c is any constant (any real numbers such as 3, 7, e, 2.5, etc.) is 0. Correspondingly, the derivative of any constant is 0. Let's take another example, the slope of a line — such as
\(y=2x\)
— is the leading coefficient, such that
\(y=2x\)
has a slope of
\(2\)
. Correspondingly, the derivatives of
\(2x\)
is
\(2\)
.
Yet what happens when functions get increasingly complicated? How can we determine the derivative? Thankfully, specific rules have been derived that govern the relationship between functions and their derivatives. As a result, for example, we are able to conclude that the slope of the tangent line of
\({x}^{2}\)
is always
\(2x\)
, and for
\(\sin{x}\)
, it is always
\(\cos{x}\)
, and so on.
Now, let us look at an important rule in differential calculus: the chain rule.

Applying the Chain Rule for Derivatives

The rule states that
\(f'(g(x))=f'(g(x))g'(x)\)
This allows us to correctly compute the derivatives of functions such as
\(\frac{d}{dx} \sin{2x}\)
.
First, we need to break the equation into components, and assign them to f and g correctly:
Let
\(f(g(x))=\sin{(g(x))}\)
and
\(g(x)={x}^{2}\)
.
The above was the most important step. Now, we just need to find
\(f'(g(x))g'(x)\)
. From other rules that we will not go into detail here, we know that the derivative of
\(\sin{g}\)
is
\(\cos{g}\)
, and the derivative of
\({x}^{2}\)
is
\(2x\)
. This allows us to use the chain rule to yield:
\(f'(g(x))g'(x)=\cos{({x}^{2})}(2x)=2x\cos{({x}^{2})}\)

Calculus Chain Rule Problems

Ready to give it a try? Tackle some of our practice calculus problems at the top of this page using the derivative chain rule, and see if you can find the answer. If you run into trouble, check out the step-by-step solution to see how the chain rule, power rule and constant factor rule can all be used together to find the derivative.

What's Next

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