Question

Calculus Question
Need to see if solution is wrong
Can someone do this and tell me if they get 63 because that's what I get however the answer says its 78???

Answer 1

13d

Answer 2

I'm getting 78

Answer 3

Can you show how please?

Answer 4

\[\begin{align*}
y&=u^3-5(u^3-7u)^2\\\\
\frac{dy}{dx}&=3u^2\frac{du}{dx}-10\left(u^3-7u\right)\left(3u^2\frac{du}{dx}-7\frac{du}{dx}\right)
\end{align*}\]
Since \(u=\sqrt x\), you have \(\dfrac{du}{dx}=\dfrac{1}{2\sqrt x}\).
\[\begin{align*}
\frac{dy}{dx}&=3\left(\sqrt x\right)^2\left(\frac{1}{2\sqrt x}\right)-10\left((\sqrt x)^3-7\sqrt x\right)\left(3(\sqrt x)^2\left(\frac{1}{2\sqrt x}\right)-\frac{7}{2\sqrt x}\right)\\\\

\frac{dy}{dx}&=\frac{3x}{2\sqrt x}-10\sqrt x\left(x-7\right)\left(\frac{3x-7}{2\sqrt x}\right)
\end{align*}\]
Now plug in \(x=4\).

Answer 5

why is it 3u^2 -10 in the derivative

Answer 6

Some simplifying first:
\[\frac{dy}{dx}=\frac{3}{2}\sqrt x-5(x-7)(3x-7)\]

Answer 7

ooh

Answer 8

I thought it was (u^3 -5)(u^3 - 7u)^2

Answer 9

I makes so much sense now
Thanks!

Answer 10

yw