# Power Inside Trigonometry Derivative with Chain Rule [closed]

Trying to differentiate the following function with the chain rule, but I'm stumped with the $$\cos(2x+3)^3$$. Do I power the $$(2x+3)$$ by $$3$$?

$${\sinh ^2 (\cos(2x+3)^3)}$$

• Take $u=2x+3$, $v=t^3$, $w=\cos(s)$, $y=\sinh(k)$ and $z=X^2$. Commented Jan 29, 2023 at 7:15
• Think that you have the structure: $f(g(h))$ with $h:=2x+3$, $g:=\cos x^3$ and $f:=\sinh^{2}x$. Then, try to write the chain rule. Commented Jan 29, 2023 at 7:23

\begin{align*} \frac{d}{dx} \left(\sinh^2 \left(\cos \left(2x+3\right)^3\right)\right) &= 2 \sinh \left(\cos \left(2x+3\right)^3\right) \cdot \frac{d}{dx}\left(\sinh \left(\cos \left(2x+3\right)^3\right)\right) \\ &= 2 \sinh \left(\cos \left(2x+3\right)^3\right) \cosh \left(\cos \left(2x+3\right)^3\right) \cdot \frac{d}{dx}\left(\cos \left(2x+3\right)^3\right). \end{align*} Can you continue from here?

As for the power, $$\cos (2x+3)^3 = \cos \left((2x+3)^3\right)$$, while $$\cos^3 (2x+3) = \left(\cos (2x+3)\right)^3$$.

As requested: \begin{align*} \frac{d}{dx}\left(\cos \left(2x+3\right)^3\right) &= -\sin(2x+3)^3 \cdot \frac{d}{dx} \left((2x+3)^3\right) \\ &= -\sin(2x+3)^3 \cdot 3(2x+3)^2 \cdot \frac{d}{dx}\left(2x+3\right) \\ &= -\sin(2x+3)^3 \cdot 3(2x+3)^2 \cdot 2. \end{align*}

• Thank you very much! I'm sorry to say that I'm still quite confused with derivative the cos (2x+3)^3. I got it correct so far until the derivative of the (cos(2x+3)^3. Can you explain how to derivate it briefly? Commented Jan 29, 2023 at 7:37
• @mcmong23 See the edit. Commented Jan 29, 2023 at 7:41
• Thank you! This really helped me understand the concept fully. Commented Jan 29, 2023 at 12:44

There is an alternative way, using the double angle formula for $$\sinh^2 x$$.

Let $$w = \cos (2x+3)^3$$. Using the identity $$\sinh^2 x = \frac {1}{2} (\cosh 2x -1)$$ and differentiating, we get $$\frac {1}{2} (\sinh 2w (2)) = \sinh 2w \ dw$$

Now we need to find $$dw$$, meaning we differentiate $$\cos (2x+3)^3$$. Letting $$z = 2x + 3$$ and $$dz = 2$$ we have $$(\cos^3 z) = -3 \cos^2 z (\sin z) (2)$$ or $$3 \sin 2z \cos z$$ when we factor out $$\cos z$$ and see that $$2 \sin z \cos z = \sin 2z$$.

Putting all the pieces together we get

$$\dfrac {dy}{dx}{\sinh ^2 (\cos(2x+3)^3)} = -\sinh 2(2x+3)\cdot 3 \sin 2(2x+3) \cos (2x+3).$$

NOTE: If you didn't want to use the double angle formulas, you can differentiate $$\sinh^2 w$$ as $$2 \sinh w \cosh w$$ and (after differentiating the other parts) leave the answer in powers of cosine and sine. The answer would then be $$\dfrac {dy}{dx}{\sinh ^2 (\cos(2x+3)^3)} = -2 \sinh (2x+3) \cosh (2x+3) \cdot 3 \sin (2x+3) \cos^2 (2x+3)$$