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Magazine Chapter 2: Problem 56
Find the derivative of the function. $$ y=e^{\cos x^{2}} \tan \left(e^{2 x}+x\right) $$
Short Answer
Expert verified
The derivative of the function is:
$$
y' = e^{\cos x^{2}}(-2x \sin x^{2}) \tan\left(e^{2x} + x\right) + e^{\cos x^{2}} \sec^2\left(e^{2x} + x\right) (2e^{2x} + 1)
$$
Step by step solution
01
Identify u(x) and v(x)
Let's consider u(x) and v(x) as:
$$
u(x) = e^{\cos x^{2}}
$$
and
$$
v(x) = \tan\left(e^{2x} + x\right)
$$
Now, we will find the derivatives for both of the chosen functions.
02
Find u'(x)
Using the chain rule to find the derivative of \(u(x)\), we have:
$$
u'(x) = \frac{d}{dx} e^{\cos x^{2}} = e^{\cos x^{2}} \frac{d}{dx} (\cos x^{2})
$$
Now, we need to find the derivative of \(\cos x^{2}\):
$$
\frac{d}{dx} (\cos x^{2}) = -\sin x^{2} \cdot 2x
$$
So, the derivative of \(u(x)\) is:
$$
u'(x) = e^{\cos x^{2}} (-\sin x^{2} \cdot 2x)
$$
03
Find v'(x)
Again, we need to use the chain rule to find the derivative of \(v(x)\), we have:
$$
v'(x) = \frac{d}{dx} \tan\left(e^{2x} + x\right) = \sec^2\left(e^{2x} + x\right) \cdot \frac{d}{dx} (e^{2x} + x)
$$
Now, we need to find the derivative of \(e^{2x} + x\):
$$
\frac{d}{dx} (e^{2x} + x) = 2e^{2x} + 1
$$
So, the derivative of \(v(x)\) is:
$$
v'(x) = \sec^2\left(e^{2x} + x\right) \cdot (2e^{2x} + 1)
$$
04
Apply the Product Rule
Now, we can apply the product rule to find the derivative of y:
\[
\begin{aligned}
y' &= u'(x) v(x) + u(x) v'(x) \\
&= {\left[ e^{\cos x^{2}} \left( -2x \sin x^{2} \right ) \right]} {\left[ \tan\left(e^{2x} + x\right) \right]} + {\left[ e^{\cos x^{2}} \right]} {\left[ \sec^2\left(e^{2x} + x\right) \cdot (2e^{2x} + 1) \right]} \\
&= e^{\cos x^{2}}(-2x \sin x^{2}) \tan\left(e^{2x} + x\right) + e^{\cos x^{2}} \sec^2\left(e^{2x} + x\right) (2e^{2x} + 1)
\end{aligned}
\]
So, the derivative of the function is:
$$
y' = e^{\cos x^{2}}(-2x \sin x^{2}) \tan\left(e^{2x} + x\right) + e^{\cos x^{2}} \sec^2\left(e^{2x} + x\right) (2e^{2x} + 1)
$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental tool in calculus for finding the derivative of a composite function. Think of it as unwrapping the layers of a function one at a time, just like peeling an onion. When a function is the result of another function nested inside it, you must use the chain rule to differentiate it. For example, if you have a function like \( f(g(x)) \), the chain rule tells us that the derivative, \( f'(g(x)) \), is the derivative of the outer function evaluated at the inner function, \( f'(g(x)) \), times the derivative of the inner function, \( g'(x) \). In our exercise, when finding the derivative of \( u(x) = e^{\cos x^{2}} \), the chain rule was employed to first differentiate the outer exponential function and then multiply it by the derivative of the inner cosine function.
Product Rule
When you're dealing with the multiplication of two functions, the product rule is your go-to technique for differentiation. Suppose you have two functions, \( u(x) \) and \( v(x) \), and you want to find the derivative of their product, \( u(x)v(x) \). According to the product rule, you don't just take the derivative of each and multiply them together. Instead, the derivative \( (uv)' \) is \( u'(x)v(x) + u(x)v'(x) \). So, you take the derivative of the first function and multiply it by the second function as it is, and then add the product of the first function as it is with the derivative of the second function. A helpful mnemonic is 鈥渇irst times the derivative of the second plus the second times the derivative of the first.鈥� The step 4 from our exercise nicely demonstrates this concept where both \( u(x) \) and \( v(x) \) and their derivatives are used to calculate the derivative of their product.
Trigonometric Functions Derivative
Trigonometric functions such as sine, cosine, and tangent are pivotal in various areas of math and science, and knowing how to differentiate them is a must for any calculus student. The derivatives of these functions follow specific rules. For sine, the derivative is cosine, and for cosine, the derivative is the negative of sine. For tangent, we have a slightly more complex rule: the derivative is secant squared. These rules allow us to tackle functions involving trigonometric terms much more confidently. In our solved problem, the presence of \( \tan(e^{2x} + x) \) required the use of the derivative of the tangent function to move towards the solution.
Exponential Functions Derivative
Exponential functions exhibit a constant rate of growth or decay and their derivatives are conveniently expressed in terms of themselves. For any exponential function of the form \( e^{f(x)} \), the derivative is \( e^{f(x)} \) multiplied by the derivative of the exponent, \( f'(x) \). This unique property where the derivative resembles the function itself makes them very easy to work with in calculus. In the step-by-step solution, when finding the derivative of \( u(x) \) and \( v(x) \), we applied this rule 鈥� thus highlighting the importance of understanding the derivative of exponential functions to solve calculus problems efficiently.
