= Solution
{c}
Write the <first strain invariant> in components of the <deformation gradient>:
$$
I_1=\operatorname{tr}(\mathbf F\mathbf F^T)
=\sum_{i,I}F_{iI}^2.
$$
Here lower-case and upper-case indices refer to the <current configuration> and <reference configuration>, respectively. Differentiating gives $\partial I_1/\partial F_{iI}=2F_{iI}$. The <chain rule> therefore gives
$$
\frac{\partial W}{\partial F_{iI}}=2w'(I_1)F_{iI},
\qquad
\partial_{\mathbf F}W=2w'(I_1)\mathbf F.
$$
Substitution into the given <hyperelastic material> law, with its indicated <matrix transpose>, yields
$$
\boxed{\mathbf T
=\mathbf F(\partial_{\mathbf F}W)^T-p\mathbf1
=2w'(I_1)\mathbf F\mathbf F^T-p\mathbf1
=2\frac{\partial w}{\partial I_1}\mathbf B-p\mathbf1.}
$$
The <left Cauchy-Green deformation tensor> is $\mathbf B=\mathbf F\mathbf F^T$. The scalar $p$ is the <Lagrange multiplier> enforcing <incompressibility>, rather than a function fixed by the <strain energy density> alone. The <identity matrix> $\mathbf1$ makes this an isotropic <constraint reaction in hyperelastic stress>.
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