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Chapter 3: Problem 10

Find the derivative of the following functions. $$g(w)=e^{w}\left(5 w^{2}+3 w+1\right)$$

Short Answer

Expert verified
Answer: The derivative of the given function is: $$\frac{dh(w)}{dw} = e^w\left(5w^2 + 13w + 4\right)$$

Step by step solution

01

Find the derivatives of the individual functions

To find the derivative of the function \(f(w)=e^w\), we differentiate the exponential function, which is: $$\frac{d}{dw}(e^w)=e^w$$ Now, to find the derivative of the function \(g(w)=5w^2 + 3w + 1\), we differentiate each term: $$\frac{d}{dw}(5w^2) = 10w$$ $$\frac{d}{dw}(3w) = 3$$ $$\frac{d}{dw}(1) = 0$$ The derivative of \(g(w)\) is the sum of these individual derivatives: $$\frac{dg(w)}{dw} = 10w + 3$$
02

Apply the product rule to find the derivative of the given function

Now we'll apply the product rule to find the derivative of the given function. The product rule states: $$(f(w)g(w))' = f'(w)g(w) + f(w)g'(w)$$ With \(f(w) = e^w\), \(f'(w) = e^w\), \(g(w) = 5w^2 + 3w + 1\), and \(g'(w) = 10w + 3\), we can substitute these functions and their derivatives into the product rule: $$\frac{d}{dw}(e^w(5w^2 + 3w + 1)) = e^w(5w^2 + 3w + 1) + e^w(10w + 3)$$
03

Simplify the expression

We can factor out \(e^w\) to combine the two terms and simplify the expression: $$\frac{d}{dw}(e^w(5w^2 + 3w + 1)) = e^w\left((5w^2 + 3w + 1) + (10w + 3)\right)$$ Now we can combine the like terms inside the parentheses: $$\frac{d}{dw}(e^w(5w^2 + 3w + 1)) = e^w\left(5w^2 + 13w + 4\right)$$ Thus, the derivative of the given function is: $$\frac{dg(w)}{dw} = e^w\left(5w^2 + 13w + 4\right)$$

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Most popular questions from this chapter

A challenging derivative Find \(\frac{d y}{d x},\) where \(\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}+x\right)=8 x y^{2}\).

One of the Leibniz Rules One of several Leibniz Rules in calculus deals with higher-order derivatives of products. Let \((f g)^{(n)}\) denote the \(n\) th derivative of the product \(f g,\) for \(n \geq 1\) a. Prove that \((f g)^{(2)}=f^{\prime \prime} g+2 f^{\prime} g^{\prime}+f g^{\prime \prime}\) b. Prove that, in general,$$(f g)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\\k\end{array}\right) f^{(k)} g^{(n-k)}$$ where \(\left(\begin{array}{l}n \\\ k\end{array}\right)=\frac{n !}{k !(n-k) !}\) are the binomial coefficients. c. Compare the result of (b) to the expansion of \((a+b)^{n}\).

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