Differentiate $y=\cosh^{3} 4x$.

Question

Differentiate $y=\cosh^{3} 4x$.

$$\frac{dy}{dx} = 3 \cosh^{2} (4x) \sinh (4x)\cdot 4$$

These are the parts that I don't quite understand:

\begin{align*} \frac {dy}{dx} &=12 \cosh^{2} (4x)\sinh (4x) \\ &=12 \cosh(4x)\cosh (4x) \sinh(4x) \\ &=12 \cosh(4x)(2 \sinh (8x))\\ &=24 \sinh (8x) \cosh (4x) \end{align*}

My questions:

• How is it that $12 \cosh^{2} (4x)\sinh (4x)$ is changed to $12 \cosh (4x) \cosh (4x)\sinh (4x) $?

• How is it that $\sinh (4x)= 2 \sinh(8x)$?

Answer 1

The first question is because $y^2 = y \cdot y$, by definition.

---

The final answer is not correct. To make it correct, we can use the identity $$\sinh 2y = 2 \sinh y \cosh y$$ Then we'll have

$$(12 \cosh(4x)) \cosh(4x) \sinh(4x) = 12\cosh 4x \left(\frac 1 2 \sinh 8x\right) = 6 \cosh 4x \sinh 8x$$

Answer 2

First of all, note that $$12\cosh (4x)\cosh (4x)\sinh(4x)\not= 12\cosh(4x)(2\sinh(8x)).$$

The following is correct :

$$\begin{align}12\cosh (4x)\cosh (4x)\sinh(4x)&= 6\cosh(4x)\times 2\cosh(4x)\sinh(4x)\\&=6\cosh(4x)\times \sinh(8x)\end{align}$$

The answer for the first question : Because $\cosh^2(x)=\cosh(x)\cosh(x).$

The answer for the second question : No, it's not true. The answer is what I wrote above.