Solveeit Logo
Login

Question

Mathematics Question on Continuity and differentiability

y=∣f(x)g(x)h(x) lmn abc∣y=\begin{vmatrix} f(x)&g(x) &h(x) \\\ l&m &n \\\ a&b &c \end{vmatrix},prove that dydx=∣f′(x)g′(x)h′(x) lmn abc∣\frac{dy}{dx}=\begin{vmatrix} f'(x)&g'(x) &h'(x) \\\ l&m &n \\\ a&b &c \end{vmatrix}

Answer

y=∣f(x)g(x)h(x) lmn abc∣y=\begin{vmatrix} f(x)&g(x) &h(x) \\\ l&m &n \\\ a&b &c \end{vmatrix}
⇒ y=(mc-nb)f(x)-(lc-na)g(x)+(lb-ma)h(x)
Then,dydx\frac{dy}{dx}=ddx\frac{d}{dx}[(mc-nb)f(x)]-ddx\frac{d}{dx}(lc-na)g(x)]+ddx\frac{d}{dx}[(lb-ma)h(x)]
=(mc-nb)f'(x)-(lc-na)g'(x)+(lb-ma)h'(x)
=∣f′(x)g′(x)h′(x) lmn abc∣\begin{vmatrix} f'(x)&g'(x) &h'(x) \\\ l&m &n \\\ a&b &c \end{vmatrix}
Thus, dydx\frac{dy}{dx}=∣f(x)g(x)h(x) lmn abc∣\begin{vmatrix} f(x)&g(x) &h(x) \\\ l&m &n \\\ a&b &c \end{vmatrix}