MHB Solving Chain Rule Derivatives: y=a^3+cos^3 (x) & y=[x+(x+sin^2 (x))^7]^5

[tc903]

My first question states y=a^3+cos^3 (x) (I couldn't quite figure out latex again.)

The derivative using the chain rule I found to be
y'=(a^3)(ln a)+3((cos (x))^2)(-sin(x)) = y' = (a^3)(ln a)-3((cos (x))^2)(sin (x))

The second question y=[x+(x+sin^2 (x))^7]^5

Derived using chain rule,
y'=5[x+(x+sin^2 (x))^7][1+7(x+sin^2 (x))^6](1+2sin (x))(cos(x))

The title wasn't meant for the latex, I have to plug these into a computer as answers. I was just wondering if I had problems with the way my answer ended or was I to continue with substituting half angle formula's such as sin^2 (x) and cos^2 (x).

 

[Ackbach]

My first question states y=a^3+cos^3 (x) (I couldn't quite figure out latex again.)

The derivative using the chain rule I found to be
y'=(a^3)(ln a)+3((cos (x))^2)(-sin(x)) = y' = (a^3)(ln a)-3((cos (x))^2)(sin (x))

If you're taking the derivative with respect to $x$, then you should leave off the logarithm term - it's a constant. I get
$y'=-3 \cos^2(x) \, \sin(x).$

The second question y=[x+(x+sin^2 (x))^7]^5

Derived using chain rule,
y'=5[x+(x+sin^2 (x))^7][1+7(x+sin^2 (x))^6](1+2sin (x))(cos(x))

Hmm, this one's a bit trickier. I get
\begin{align*}
y&=\left[x+(x+\sin^2(x))^7\right]^5 \\
y'&=5\left[x+(x+\sin^2(x))^7\right]^4 \cdot \left[1+7(x+\sin^2(x))^6(1+2\sin(x)\cos(x))\right].
\end{align*}

The title wasn't meant for the latex, I have to plug these into a computer as answers. I was just wondering if I had problems with the way my answer ended or was I to continue with substituting half angle formula's such as sin^2 (x) and cos^2 (x).

I don't know that I would bother about half-angle formulas. But I would give you this advice: almost no computer system understands cos^2(x). You should write it as (cos(x))^2.

Does this answer your question?