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Recall that:
[tex]\frac{d}{du}\sinh (u)=\cosh (u).[/tex]Applying the chain rule we get that:
[tex]\frac{d}{dx}\sinh (x^5)=\frac{d}{du}\sinh (u)|_{u=x^5}\cdot\frac{d}{dx}(x^5)\text{.}[/tex]Therefore:
[tex]\frac{d}{dx}\sinh (x^5)=\cosh (u)|_{u=x^5}\cdot5x^{5-1}\text{.}[/tex]Simplifying the above result we get:
[tex]\begin{gathered} \frac{d}{dx}\sinh (x^5)=\cosh (x^5)\cdot5x^4 \\ =5x^4\cosh (x^5)\text{.} \end{gathered}[/tex]Answer:
[tex]h^{\prime}(x)=5x^4\cosh (x^5)\text{.}[/tex]