Question

$$\left. \begin{array} { l } { \theta _ { 1 } ( e ^ { 8 } - x + i ) ^ { 6 } } \\ { y = ( 2 x + 1 ) ^ { 7 } ( e ^ { 8 } - x + i ) ^ { 6 } } \end{array} \right.$$

Answer

$$r=((2*x+1)^7*(x^3-x+1)^6)/(Di*e^2*IM*f^2*n*t^2*a*o*y)$$

Solution


Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[Di{f}^{2}{e}^{2}rn{t}^{2}\imath aoy={(2x+1)}^{7}{({x}^{3}-x+1)}^{6}\]
Regroup terms.
\[Di{e}^{2}\imath {f}^{2}rn{t}^{2}aoy={(2x+1)}^{7}{({x}^{3}-x+1)}^{6}\]
Divide both sides by \(Di\).
\[{e}^{2}\imath {f}^{2}rn{t}^{2}aoy=\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di}\]
Divide both sides by \({e}^{2}\).
\[\imath {f}^{2}rn{t}^{2}aoy=\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di}}{{e}^{2}}\]
Simplify  \(\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di}}{{e}^{2}}\)  to  \(\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}}\).
\[\imath {f}^{2}rn{t}^{2}aoy=\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}}\]
Divide both sides by \(\imath \).
\[{f}^{2}rn{t}^{2}aoy=\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}}}{\imath }\]
Simplify  \(\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}}}{\imath }\)  to  \(\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath }\).
\[{f}^{2}rn{t}^{2}aoy=\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath }\]
Divide both sides by \({f}^{2}\).
\[rn{t}^{2}aoy=\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath }}{{f}^{2}}\]
Simplify  \(\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath }}{{f}^{2}}\)  to  \(\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}}\).
\[rn{t}^{2}aoy=\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}}\]
Divide both sides by \(n\).
\[r{t}^{2}aoy=\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}}}{n}\]
Simplify  \(\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}}}{n}\)  to  \(\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n}\).
\[r{t}^{2}aoy=\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n}\]
Divide both sides by \({t}^{2}\).
\[raoy=\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n}}{{t}^{2}}\]
Simplify  \(\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n}}{{t}^{2}}\)  to  \(\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}}\).
\[raoy=\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}}\]
Divide both sides by \(a\).
\[roy=\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}}}{a}\]
Simplify  \(\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}}}{a}\)  to  \(\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}a}\).
\[roy=\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}a}\]
Divide both sides by \(o\).
\[ry=\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}a}}{o}\]
Simplify  \(\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}a}}{o}\)  to  \(\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}ao}\).
\[ry=\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}ao}\]
Divide both sides by \(y\).
\[r=\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}ao}}{y}\]
Simplify  \(\frac{\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}ao}}{y}\)  to  \(\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}aoy}\).
\[r=\frac{{(2x+1)}^{7}{({x}^{3}-x+1)}^{6}}{Di{e}^{2}\imath {f}^{2}n{t}^{2}aoy}\]