Question

for f(x) = 4x^4, find dy/dx using the definition, dy/dx = lim(δx→0) δy/δx = lim(δx→0) f(x + δx) - f(x)/δx. (a) first, evaluate δy = f(x + δx) - f(x) and express it in the form δy = l(x,δx)·δx. use dx to represent δx. δy = δx (b) using the l(x,δx)δx above, find the simplified derivative dy/dx = lim(δx→0) δy/δx. dy/dx = lim(δx→0) δy/δx = note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining. email instructor

Explanation:

Step1: Find $\Delta y$

First, find $f(x+\Delta x)$ where $f(x) = 4x^{4}$. So $f(x+\Delta x)=4(x + \Delta x)^{4}$.
Using the binomial expansion $(a + b)^{4}=a^{4}+4a^{3}b + 6a^{2}b^{2}+4ab^{3}+b^{4}$, with $a = x$ and $b=\Delta x$, we have $f(x+\Delta x)=4(x^{4}+4x^{3}\Delta x + 6x^{2}(\Delta x)^{2}+4x(\Delta x)^{3}+(\Delta x)^{4})=4x^{4}+16x^{3}\Delta x + 24x^{2}(\Delta x)^{2}+16x(\Delta x)^{3}+4(\Delta x)^{4}$.
Then $\Delta y=f(x+\Delta x)-f(x)=(4x^{4}+16x^{3}\Delta x + 24x^{2}(\Delta x)^{2}+16x(\Delta x)^{3}+4(\Delta x)^{4})-4x^{4}=16x^{3}\Delta x + 24x^{2}(\Delta x)^{2}+16x(\Delta x)^{3}+4(\Delta x)^{4}=(16x^{3}+24x^{2}\Delta x + 16x(\Delta x)^{2}+4(\Delta x)^{3})\Delta x$.

Step2: Find the derivative

We know that $\frac{dy}{dx}=\lim_{\Delta x
ightarrow0}\frac{\Delta y}{\Delta x}$. Since $\Delta y=(16x^{3}+24x^{2}\Delta x + 16x(\Delta x)^{2}+4(\Delta x)^{3})\Delta x$, then $\frac{\Delta y}{\Delta x}=16x^{3}+24x^{2}\Delta x + 16x(\Delta x)^{2}+4(\Delta x)^{3}$.
Taking the limit as $\Delta x
ightarrow0$, we have $\lim_{\Delta x
ightarrow0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x
ightarrow0}(16x^{3}+24x^{2}\Delta x + 16x(\Delta x)^{2}+4(\Delta x)^{3}) = 16x^{3}$.

Answer:

(a) $(16x^{3}+24x^{2}\Delta x + 16x(\Delta x)^{2}+4(\Delta x)^{3})\Delta x$
(b) $16x^{3}$