Question

Use $x=e^y$ to explain why $\frac{d}{dx}(\ln x)=\frac{1}{x},$ for $x>0$.

Short answer

Question: Use the function $x=e^y$ to explain why $\frac{d}{dx}(\ln x)=\frac{1}{x}$ for $x>0$. Answer: By differentiating the function $x=e^y$ with respect to $y$, expressing $y$ in terms of $x$, and using the Chain Rule, we were able to show that $\frac{d}{dx}(\ln x)=\frac{1}{x}$ for $x>0$.

Step-by-step solution

Differentiate the given function with respect to y

Differentiate the function $x=e^y$ with respect to $y$. To do this, we need to find the derivative of $e^y$, which is simply $e^y$. So, the derivative with respect to $y$ is: $$\frac{dx}{dy}=e^y$$

Express y in terms of x

We need to express $y$ in terms of $x$ to switch from $dy$ to $dx$. From the given function $x=e^y$, we can find $y$ as follows: $$y=\ln x$$

Use the Chain Rule to differentiate $\ln x$ with respect to x

To find the derivative of $\ln x$ with respect to $x$, we can use the Chain Rule. The chain rule states that: $$\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}$$ In our case, $u=\ln x$, and from Step 1, we found that $dx=e^{y}dy$. Thus, we can find $\frac{du}{dx}$ as follows: $$\frac{d}{dx}(\ln x)=\frac{1}{e^y}\cdot \frac{d}{dy}(e^y)$$

Substitute y in terms of x

We know from Step 2 that $y=\ln x$. We can use this to express all the terms in the previous step in terms of $x$. Using the relationship between $y$ and $x$, we have: $$\frac{d}{dx}(\ln x)=\frac{1}{e^{\ln x}}\cdot e^{\ln x}$$

Simplify the result

Finally, we know that $e^{\ln x}=x$. We can use this to simplify the expression we found in the previous step: $$\frac{d}{dx}(\ln x)=\frac{1}{x}\cdot x=\frac{1}{x}$$ Therefore, we have shown that $\frac{d}{dx}(\ln x)=\frac{1}{x}$ for $x>0$.

Key concepts

Chain Rule in Calculus

Understanding the chain rule is like unlocking a crucial calculus treasure chest – it lets you tackle derivatives of composite functions with ease. Imagine you have a function that's the combination of two other functions, like a Russian doll. You want to differentiate this nested function, right? The chain rule is your go-to tool.

Let's say we have a function of a function: like speed as a function of time, within time as a function of distance. The chain rule tells us how to find the derivative of this combo. The rule says: take the derivative of the outer function, but leave the inner function untouched. Then multiply that by the derivative of the inner function.

In mathematical terms, for functions $u(x)$ defined as another function $g(f(x))$, the derivative of $u$ with respect to $x$ is $u^{\prime }(x)=g^{\prime }(f(x))\cdot f^{\prime }(x)$. In the given exercise, this principle allows us to connect changes in $x$ with changes in $y$, even when $y$ itself is a function of $x$ through the natural logarithm.

Exponential Functions

Exponential functions, such as $e^x$, are the mathematical equivalent of growth superstars—they model scenarios like continuously compounding interest or populations growing without any limits. The base $e$ is an irrational number, approximately equal to 2.71828, and it's known as Euler's number.

The beauty of the exponential function lies in its simplicity when it comes to differentiation. The derivative of $e^x$ with respect to $x$ is simply $e^x$ itself. This unique property makes $e^x$ a fundamental example when learning calculus because it's so predictable and forms the backbone for many more complex equtions. In our exercise, we use this property to seamlessly transition between expressions involving $y$ and $x$, thanks to the relationship $x=e^y$, which makes differentiation a walk in the park.

Natural Logarithm Properties

The natural logarithm, represented as $\ln (x)$, is a special creature in the calculus zoo. It's the inverse function of the exponential function $e^x$. This nifty partnership means that applying $\ln$ to $e^x$ (or vice versa) is like doing a forward flip followed by a backward flip—it takes you back to where you started.

Some cool properties of the natural logarithm include $\ln (1)=0$ and $\ln (e)=1$. One particularly relevant property for our exercise is that $e^{\ln (x)}=x$, and this is essential for simplifying derivatives involving the natural logarithm. It's this cancellation property that allows us to neatly tie $y$ back to $x$ when differentiating, thereby simplifying the derivative of $\ln (x)$ to $1/x$.

Implicit Differentiation

Sometimes, you'll come across a function that's so intertwined, it feels like untangling a pair of earphones that've been in your pocket for ages. Implicit differentiation is the tool you need for just such a tangle. It allows us to find the derivative of a function where the variables are mixed together, like chocolate chips in cookie dough.

It's not always possible (or easy) to separate the variables neatly on either side of the equation. Implicit differentiation steps in to save the day by differentiating each side of the equation with respect to the independent variable, while treating the other variables as functions of that independent variable.

In the exercise, we use implicit differentiation to take the derivative of $x$ in terms of $y$, without the need to solve explicitly for $y$ first. It's a more efficient path to arrive at the derivative of $\ln (x)$, demonstrating how implicit differentiation is a versatile technique in the world of calculus.