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Differentiate the following w.r.t x: $\sin^n(ax^2+bx+c)$.

$\begin{array}{1 1} (A)\;n(2ax+b)\sin^{n-1}(ax^2+bx+c).\cos(ax^2+bx+c) \\(B)\;n(2ax+b)\sin^{n}(ax^2+bx+c).\cos(ax^2+bx+c) \\ (C)\;-n(2ax+b)\sin^{n-1}(ax^2+bx+c).\cos(ax^2+bx+c) \\(D)\;-n(2ax+b)\sin^{n}(ax^2+bx+c).\cos(ax^2+bx+c) \end{array} $

1 Answer

Toolbox:
  • Chain Rule : Suppose a real valued function which is a composite of two function $u$ and $v$ $(i.e$) $f=uov$ then $\large\frac{df}{dx}=\frac{dv}{dt}$$\times \large\frac{dt}{dx}$
Step 1:
$y=\sin^n(ax^2+bx+c)$
Let $u=ax^2+bx+c$
$\large\frac{du}{dx}$$=2ax+b$
Hence $y=\sin^nu$
$\large\frac{dy}{du}$$=n\sin^{n-1}u.\cos u$
Therefore $\large\frac{dy}{dx}=\frac{dy}{du}$$\times \large\frac{du}{dx}$
$\qquad\qquad\;\;\;\;=n\sin^{n-1}u\cos u\times (2ax+b)$
Step 2:
Substituting for $u$ we get,
$\large\frac{dy}{dx}$$=[n\sin^{n-1}(ax^2+bx+c).\cos(ax^2+bx+c)]\times (2ax+b)$
$\quad\;=n(2ax+b)\sin^{n-1}(ax^2+bx+c).\cos(ax^2+bx+c)$
answered Jun 27, 2013 by sreemathi.v
 
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