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Find the derivative of the following functions. $$y=5 \cdot 4^{x}$$

Short Answer

Expert verified
Answer: The derivative of the function \(y = 5 \cdot 4^{x}\) with respect to x is \(\frac{dy}{dx} = 5\ln(4) \cdot 4^{x}\).

Step by step solution

01

Identify the function

The given function is an exponential function: $$y = 5 \cdot 4^{x}$$
02

Apply the chain rule for differentiation

The chain rule states that: $$(f(g(x)))' = f'(g(x)) \cdot g'(x)$$ Here, we can rewrite the given function as follows with \(f(g(x))\): $$f(u) = 5u \space \text{where} \space u = 4^{x}$$
03

Differentiate the inner function with respect to x

Differentiate the inner function \(u\) with respect to \(x\): $$u = 4^{x}$$ So its derivative \(u'\) can be found as: $$u' = \frac{d(4^{x})}{dx} = 4^{x} \cdot \ln(4)$$
04

Differentiate the outer function with respect to u

Next, differentiate the outer function \(f(u)\) with respect to \(u\): $$f(u) = 5u$$ So its derivative \(f'\) can be found as: $$f'(u) = \frac{d(5u)}{du} = 5$$
05

Apply the chain rule

Now, using the chain rule, we can find the derivative of the given function with respect to x: $$\frac{dy}{dx} = f'(u) \cdot u'(x)$$ Plug in the values of \(f'(u)\) and \(u'(x)\) $$\frac{dy}{dx} =5 \cdot (4^{x} \cdot \ln(4))$$
06

Simplify the derivative function

Simplify the derivative of the function: $$\frac{dy}{dx} = 5\ln(4) \cdot 4^{x}$$ Now, we have found the derivative of the given function: $$\frac{dy}{dx} = 5\ln(4) \cdot 4^{x}$$

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Key Concepts

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

Chain Rule for Differentiation
The Chain Rule is a fundamental strategy used in calculus for finding the derivative of composite functions. It's like a mathematical version of a Russian nesting doll; inside each function, there is another function that also requires differentiation. When faced with the task of finding the derivative of a composite function, such as one where an exponential function is being multiplied by a constant, the Chain Rule allows us to differentiate the function in layers.

To use the Chain Rule, first, identify the inner and outer functions. Then, differentiate each function separately. You'll end up with an expression for the derivative of the outer function, and another for the inner function. The final step is to multiply these two derivatives together. If it sounds complex, don't worry, we'll go over an example below to clear things up!
Derivative of Exponential Function
Exponential functions are unique in that the rate of growth or decay is proportional to the function's current value. To differentiate an exponential function like \( y = a^{x} \), where 'a' is a constant, you need to apply a special rule. The derivative of \( a^{x} \) with respect to 'x' is \(a^{x} \times \text{ln}(a)\).

Remember that 'e', the natural exponential base, is a unique case where the derivative of \(e^{x}\) is simply \(e^{x}\) itself. For other bases, like in the given exercise where we have base 4, the natural logarithm of the base, ln(a), comes into play.
Applying the Chain Rule
Let’s put the Chain Rule into action with the example from the exercise. We have the function \( y = 5 \times 4^{x} \). Here, the '5' can be seen as the outer function, whereas \(4^{x}\) serves as the inner function.

First, we differentiate \(4^{x}\) to get \(4^{x} \times \text{ln}(4)\). Then we acknowledge that the derivative of '5' with respect to \(4^{x}\) is simply '5', since '5' is a constant and does not change with \(4^{x}\). Finally, according to the Chain Rule, we multiply the derivatives of the inner and outer functions together: \(5 \times (4^{x} \times \text{ln}(4))\).
Simplifying Derivatives
At the end of the differentiation process, simplifying the derivative can make it easier to understand and further work with. To simplify, we often combine constants or like terms. For instance, in the example \( \frac{dy}{dx} = 5 \times (4^{x} \times \text{ln}(4)) \), we can combine the constant '5' and \( \text{ln}(4) \) to get a single constant term.

Simplification may sometimes involve more than just combining terms; it can include factoring, reducing fractions, or utilizing properties of exponents and logarithms. The goal is to achieve the most concise and manageable form of the derivative for practical use.

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Most popular questions from this chapter

a. Determine an equation of the tangent line and the normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises 73-78. b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}\right)^{2}=\frac{25}{3}\left(x^{2}-y^{2}\right); \left(x_{0}, y_{0}\right)=(2,-1)\) (lemniscate of Bernoulli) (Graph cant copy)

Carry out the following steps. a. Verify that the given point lies on the curve. b. Determine an equation of the line tangent to the curve at the given point. $$x^{4}-x^{2} y+y^{4}=1 ;(-1,1)$$ (Graph cant copy)

The hands of the clock in the tower of the Houses of Parliament in London are approximately \(3 \mathrm{m}\) and \(2.5 \mathrm{m}\) in length. How fast is the distance between the tips of the hands changing at \(9.00 ?\) (Hint: Use the Law of Cosines.)

Beginning at age \(30,\) a self-employed plumber saves \(\$ 250\) per month in a retirement account until he reaches age \(65 .\) The account offers \(6 \%\) interest, compounded monthly. The balance in the account after \(t\) years is given by \(A(t)=50,000\left(1.005^{12 t}-1\right)\) a. Compute the balance in the account after \(5,15,25,\) and 35 years. What is the average rate of change in the value of the account over the intervals \([5,15],[15,25],\) and [25,35]\(?\) b. Suppose the plumber started saving at age 25 instead of age 30\. Find the balance at age 65 (after 40 years of investing). c. Use the derivative \(d A / d t\) to explain the surprising result in part (b) and the advice: Start saving for retirement as early as possible.

Robert Boyle \((1627-1691)\) found that for a given quantity of gas at a constant temperature, the pressure \(P\) (in kPa) and volume \(V\) of the gas (in \(m^{3}\) ) are accurately approximated by the equation \(V=k / P\), where \(k>0\) is constant. Suppose the volume of an expanding gas is increasing at a rate of \(0.15 \mathrm{m}^{3} / \mathrm{min}\) when the volume \(V=0.5 \mathrm{m}^{3}\) and the pressure is \(P=50 \mathrm{kPa}\). At what rate is pressure changing at this moment?

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