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Chapter 16: Problem 10

Find \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}\) given \(y=\mathrm{e}^{\left(x^{2}\right)}\).

Short Answer

Expert verified
Answer: The second derivative of the function is \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} = 2\mathrm{e}^{\left(x^{2}\right)}(1 + 4x^2)\).

Step by step solution

01

Find the first derivative

To find the first derivative, we will differentiate the function \(y=\mathrm{e}^{\left(x^{2}\right)}\) with respect to x. We'll use the chain rule, which states that if \(y(u) = \mathrm{e}^{u}\) and \(u = x^{2}\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x} = \frac{\mathrm{d} y}{\mathrm{~d} u} \cdot \frac{\mathrm{d} u}{\mathrm{~d} x}\). First, find \(\frac{\mathrm{d} y}{\mathrm{~d} u}\) by differentiating \(y=\mathrm{e}^{\left(u\right)}\): \(\frac{\mathrm{d} y}{\mathrm{~d} u} = \mathrm{e}^{u}\). Next, find \(\frac{\mathrm{d} u}{\mathrm{~d} x}\) by differentiating \(u=x^{2}\): \(\frac{\mathrm{d} u}{\mathrm{~d} x} = 2x\). Now, multiply the two results to obtain the first derivative \(\frac{\mathrm{d} y}{\mathrm{~d} x}\): \(\frac{\mathrm{d} y}{\mathrm{~d} x} = \mathrm{e}^{u} \cdot 2x = \mathrm{e}^{\left(x^{2}\right)} \cdot 2x\).
02

Find the second derivative

To find the second derivative, we will differentiate the first derivative \(\frac{\mathrm{d} y}{\mathrm{~d} x} = \mathrm{e}^{\left(x^{2}\right)} \cdot 2x\) with respect to x. Again, we'll use the chain rule. Let \(v = x^{2}\), and rewrite the first derivative as a product of two functions: \(\frac{\mathrm{d} y}{\mathrm{~d} x} = 2x \cdot \mathrm{e}^{v}\). Now we apply the product rule: \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} = \frac{\mathrm{d}}{\mathrm{~d} x}(2x) \cdot \mathrm{e}^{v} + 2x \cdot \frac{\mathrm{d}}{\mathrm{~d} x}(\mathrm{e}^{v})\). First, differentiate \(2x\) with respect to x: \(\frac{\mathrm{d}}{\mathrm{~d} x}(2x) = 2\). Now, differentiate \(\mathrm{e}^{v}\) with respect to x using the chain rule: \(\frac{\mathrm{d}}{\mathrm{~d} x}(\mathrm{e}^{v}) = \frac{\mathrm{d}}{\mathrm{~d} v}(\mathrm{e}^{v}) \cdot \frac{\mathrm{d} v}{\mathrm{~d} x} = \mathrm{e}^{v} \cdot 2x = \mathrm{e}^{\left(x^{2}\right)} \cdot 2x\). Combine the results to find the second derivative: \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} = 2 \cdot \mathrm{e}^{\left(x^{2}\right)} + 2x \cdot \left(\mathrm{e}^{\left(x^{2}\right)} \cdot 2x\right) = 2\mathrm{e}^{\left(x^{2}\right)}(1 + 4x^2)\). Thus, the second derivative of the given function is \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} = 2\mathrm{e}^{\left(x^{2}\right)}(1 + 4x^2)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
Understanding the chain rule is essential when dealing with composite functions, where one function is nested within another. In our exercise, we have an exponential function with a power of x^2. Imagine peeling an onion: The outermost layer is the exponential function, and the layer beneath is x^2.

Applying the Chain Rule


When differentiating, we tackle each 'layer' separately. Start with the outer function, treating the inner function as a constant. In this case, we differentiate e^u with respect to u, resulting in e^u, because the derivative of e^x with respect to x is e^x. Next, we differentiate the inner function u=x^2 with respect to x, yielding 2x. Multiplying these two parts gives us the derivative of the whole composite function.
Product Rule
When we differentiate products of two functions like f(x)g(x), the product rule comes to the rescue. This rule asserts that the derivative of a product is not merely the product of the derivatives. Instead, we need to apply a special formula:

Formula for the Product Rule


The product rule says that d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x), which translates to 'the derivative of the first function times the second function plus the first function times the derivative of the second function.'

In our second derivative calculation, we used the product rule on the first derivative, 2x * e^(x^2), following the aforementioned formula to differentiate this product with respect to x. The key takeaway here is to remember to differentiate each function separately, and then combine the results according to the product rule.
Exponential Function Differentiation
Differentiating exponential functions hinges on a unique property: the derivative of e^x with respect to x is again e^x. This property remains true for any exponential function with base e, the natural logarithm base.

Simplifying with Exponentials


When we differentiate exponential functions, they often simplify the process because they remain unchanged in their differentiated form. For our exercise with a function y=e^(x^2), we apply this property, and it grants us an elegant and straightforward component of e^(x^2) during differentiation using the chain rule. The behavior of exponential functions streamlines calculations and preserves the form of the function throughout differentiation steps, making it easier to work with successive derivatives like the second derivative we sought in the exercise.

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Most popular questions from this chapter

Calculate the equation of the tangent to \(y=\sin x\) where \(x=\frac{\pi}{4}\).

Find the derivative of each of the following functions: (a) \(z(t)=\mathrm{e}^{3 t}(2 t-5)^{3}(3 t+1)^{4}\) (b) \(h(t)=3 \mathrm{e}^{-6 t} t^{7}(t+6)^{33}\) (c) \(M(p)=-p^{4} \sin ^{5} p\) (d) \(P(r)=6\left(1+r^{2}\right)^{6} \sqrt{3+r^{2}}\) (e) \(y(x)=\frac{(7-x)^{3}}{\left(3 x^{2}+1\right)^{2}}\)

Find the derivative of the following functions using logarithmic differentiation: (a) \(a(t)=\left(1+t^{2}\right)^{3}\left(1-t^{2}\right)^{4} \sin ^{3} t\) (b) \(b(r)=\mathrm{e}^{-r} \sin ^{5} 2 r \cos ^{4} 3 r\) (c) \(K(p)=6 p \sqrt{\sin p}(\cos 2 p)^{1 / 3}\) (d) \(N(y)=\frac{(1+\sqrt{y})^{4}(6+7 y)^{3}}{\sqrt{1+y}}\) (e) \(x(t)=\frac{6}{(2+t)^{3}(1-t)^{4} \cos ^{3} t}\)

Find the equation of the tangent to \(y=3 x^{3}-x\) where \(x=1\)

Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) given (a) \(x^{2}+x^{3}+y^{2}-y^{3}=1\) (b) \(2 x^{2}-y^{2}+3 x y-7 x-10 y=0\) (c) \(x y^{2}+\frac{y}{x}=\mathrm{e}^{x}\) (d) \(\ln (x y)-\sqrt{x}=\sqrt{y}\) (e) \(x \sin y+y^{2} \cos 2 x=y\)
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