Question

Compute the derivative of the given function. $$g(t)=5^{\cos t}$$

Short answer

The derivative is \( g'(t) = -\sin t \cdot \ln 5 \cdot 5^{\cos t} \).

Step-by-step solution

Understand the Function

We are given the function \( g(t) = 5^{\cos t} \). This is an exponential function with respect to \( \cos t \) and a base of 5.

Apply Logarithmic Differentiation

Since the function is in the form \( a^{f(x)} \), where \( a = 5 \) and \( f(x) = \cos t \), we will use logarithmic differentiation. Start by taking the natural logarithm on both sides: \( \ln(g(t)) = \ln(5^{\cos t}) \). Using the logarithm power rule, simplify to: \( \ln(g(t)) = \cos t \cdot \ln 5 \).

Differentiate Implicitly

Differentiate both sides with respect to \( t \). For the left side, use the chain rule: \( \frac{d}{dt}[\ln(g(t))] = \frac{1}{g(t)} \cdot g'(t) \). For the right side, differentiate \( \cos t \cdot \ln 5 \) with respect to \( t \), giving: \( -\sin t \cdot \ln 5 \).

Solve for \( g'(t) \)

Equating the derivatives from Step 3: \( \frac{1}{g(t)} \cdot g'(t) = -\sin t \cdot \ln 5 \). Multiply both sides by \( g(t) \) to isolate \( g'(t) \): \( g'(t) = -\sin t \cdot \ln 5 \cdot 5^{\cos t} \).

Write the Final Derivative

The derivative of the function \( g(t) = 5^{\cos t} \) is \( g'(t) = -\sin t \cdot \ln 5 \cdot 5^{\cos t} \).

Key concepts

Logarithmic Differentiation

Logarithmic differentiation is a useful technique when differentiating functions where a variable is in the exponent, like the function we have: \( g(t) = 5^{\cos t} \). To make this process easier, we apply the natural logarithm to both sides of the equation. Taking the natural logarithm simplifies exponents considerably. In our case, applying \( \ln \) helps us bring the cosine function down in front of the logarithm: \( \ln(g(t)) = \cos t \cdot \ln 5 \). This technique is especially handy when dealing with the derivatives of exponential functions involving products, quotients, or powers. Finally, differentiating such simplified expressions becomes more straightforward compared to their original forms.

Implicit Differentiation

Implicit differentiation is a method used when we have one variable defined implicitly in terms of another, rather than explicitly. It's about differentiating both sides of an equation with respect to a variable like \( t \).
In the function \( g(t) = 5^{\cos t} \), after applying logarithmic differentiation, we essentially end up with an implicit function: \( \ln(g(t)) = \cos t \cdot \ln 5 \). We then differentiate both sides with respect to \( t \), remembering to use the chain rule as well.
For the left side, when differentiating \( \ln(g(t)) \), you apply the chain rule, which gives us \( \frac{1}{g(t)} \cdot g'(t) \). And on the right, differentiating \( \cos t \cdot \ln 5 \) is straightforward, yielding \( -\sin t \cdot \ln 5 \). This shows how implicit differentiation unravels the layers in complex functions.

Chain Rule

The chain rule is a fundamental concept required for differentiating compositions of functions. It is especially important when working with nested functions, like exponential functions where the exponent itself is a function.
In our exercise with \( g(t) = 5^{\cos t} \), the chain rule is applied when differentiating \( \ln(g(t)) \). Differentiating \( \ln(g(t)) \) requires the chain rule since \( g(t) \) itself is a function of \( t \). So, it gives us \( \frac{1}{g(t)} \cdot g'(t) \). This rule helps us find the derivative by recognizing how changes in \( t \) affect both \( \cos t \) and subsequently \( 5^{\cos t} \), through a cascade of changes impacting each part. Using the chain rule allows us to effectively peel back these layers and compute the derivative accurately.