3.6 The chain rule  (Page 2/6)

First, rewrite $h\left(x\right)=\frac{1}{{\left(3x^2+1\right)}^2}={\left(3x^2+1\right)}^{\mathrm{-2}}.$

Applying the power rule with $g\left(x\right)=3x^2+1,$ we have

Rewriting back to the original form gives us

First recall that ${\text{sin}}^{3}x={\left(\text{sin}x\right)}^3,$ so we can rewrite $h\left(x\right)={\text{sin}}^{3}x$ as $h\left(x\right)={\left(\text{sin}x\right)}^3.$

Applying the power rule with $g\left(x\right)=\text{sin}x,$ we obtain

Because we are finding an equation of a line, we need a point. The x -coordinate of the point is 2. To find the y -coordinate, substitute 2 into $h(x).$ Since $h\left(2\right)=\frac{1}{{\left(3\left(2\right)-5\right)}^2}=1,$ the point is $\left(2,1\right).$

For the slope, we need $h^{\prime }(2).$ To find $h^{\prime }(x),$ first we rewrite $h\left(x\right)={\left(3x-5\right)}^{\mathrm{-2}}$ and apply the power rule to obtain

By substituting, we have $h^{\prime }\left(2\right)=\mathrm{-6}{\left(3\left(2\right)-5\right)}^{\mathrm{-3}}=\mathrm{-6}.$ Therefore, the line has equation $y-1=\mathrm{-6}\left(x-2\right).$ Rewriting, the equation of the line is $y=\mathrm{-6}x+13.$