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Please find the derivative h(x)= sinh(x^5)

Sagot :

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]