QUESTION IMAGE
Question
write the composite function in the form ( f(g(x)) ). identify the inner function ( u = g(x) ) and the outer function ( y = f(u) ). (use non - identity functions for ( f(u) ) and ( g(x) ).) ( y=sqrt{x^{3}+6} ) ( (f(u),g(x))=(quad) ) find the derivative ( \frac{dy}{dx} ) ( \frac{dy}{dx}=quad )
Step1: Identify inner and outer functions
Let $u = g(x)=x^{3}+6$ and $f(u)=\sqrt{u}=u^{\frac{1}{2}}$. So, $(f(u),g(x))=(\sqrt{u},x^{3}+6)$.
Step2: Find derivatives of inner and outer functions
The derivative of $g(x)$ with respect to $x$ is $g^\prime(x)=\frac{d}{dx}(x^{3}+6)=3x^{2}$. The derivative of $f(u)$ with respect to $u$ is $f^\prime(u)=\frac{d}{du}(u^{\frac{1}{2}})=\frac{1}{2}u^{-\frac{1}{2}}$.
Step3: Apply chain - rule
By the chain - rule $\frac{dy}{dx}=f^\prime(g(x))\cdot g^\prime(x)$. Substitute $u = g(x)=x^{3}+6$ into $f^\prime(u)$ and multiply by $g^\prime(x)$. We get $\frac{dy}{dx}=\frac{1}{2}(x^{3}+6)^{-\frac{1}{2}}\cdot3x^{2}=\frac{3x^{2}}{2\sqrt{x^{3}+6}}$.
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$(f(u),g(x))=(\sqrt{u},x^{3}+6)$
$\frac{dy}{dx}=\frac{3x^{2}}{2\sqrt{x^{3}+6}}$