# Let z=e^(x^2+3y)+x^3y^2, where x=tcos r and y = rt^4. Use the chain rule fot multivariable functions (Cals III Chain Rule) to find (delz)/(delr) and (delz)/delt). Give your answers in terms of r and t only. Be sure to show all of your work.

Let $z={e}^{{x}^{2}+3y}+{x}^{3}{y}^{2}$, where $x=t\mathrm{cos}r\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r{t}^{4}$. Use the chain rule fot multivariable functions (Cals III Chain Rule) to find $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial }t\right)$. Give your answers in terms of r and t only. Be sure to show all of your work.
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Clelioo

$z={e}^{{x}^{2}+3y}+{x}^{3}{y}^{2}$ (1) where Now use the chain rule for multivariable function then $\left(\partial \left({e}^{{x}^{2}+3y}+{x}^{3}{y}^{3}\right)\right)/\left(\partial y\right)×\left(\partial \left(r{t}^{4}\right)\right)/\left(\partial r\right)\left(\partial z\right)/\left(\partial r\right)=\left({e}^{{x}^{2}+3y}\ast 2x+3{x}^{2}{y}^{2}\right)×\left(-tsinr\right)+\left({e}^{\left(}{x}^{2}+3y\right)\ast 2x+3{x}^{2}{y}^{2}\right)×{t}^{4}\left(\partial z\right)/\left(\partial r\right)$$=-t\mathrm{sin}r\left(2x{e}^{{x}^{2}+3y}\ast 2x+3{x}^{2}{y}^{2}\right)+{t}^{4}\left(3{e}^{{x}^{2}+3y}\ast 2x+3{x}^{3}y\right)$ Now put $r\ast {e}^{\left(t\mathrm{cos}r{\right)}^{2}+3r{t}^{4}}+3\left(t\mathrm{cos}r{\right)}^{2}\left(r{t}^{4}{\right)}^{2}\right)+{t}^{4}\left(3{e}^{\left(t\mathrm{cos}r{\right)}^{2}}+3r{t}^{4}\right)+2\left(t\mathrm{cos}r{\right)}^{3}r{t}^{4}\right)\left(\partial z\right)/\left(\partial r\right)=$$-\left(2{t}^{2}\mathrm{sin}r\mathrm{cos}r\ast {r}^{\left(t\mathrm{cos}r{\right)}^{2}}+3r{t}^{4}\right)+3{r}^{2}{t}^{11}\mathrm{sin}r{\mathrm{cos}}^{2}r\right)+\left(3{t}^{4}{e}^{\left(t\mathrm{cos}r{\right)}^{2}}+3r{t}^{4}\right)+2r{t}^{11}co{s}^{3}r\right)\left(\partial z\right)/\left(\partial r\right)=$${}^{\left(t\mathrm{cos}r{\right)}^{2}}+3r{t}^{4}\right)\left(-{t}^{2}\mathrm{sin}2r+3{t}^{4}\right)-3{r}^{2}{t}^{11}\mathrm{sin}r{\mathrm{cos}}^{2}r+2r{t}^{11}{\mathrm{cos}}^{3}r\left[\because ,\mathrm{sin}2x=2\mathrm{sin}x\mathrm{cos}x\right]$