Question

nth derivative of y =sin2x.cos3x​

Answer 1

Answer:

We can find nth derivative of $y=sin^2x.cos^3x$ using trigonometric formula.

Answer:  $y= \frac{1}{16} (2cos(\frac{n\pi }{2} +x)-5^n.cos(\frac{n\pi }{2} +5x)-3^n.cos(\frac{n\pi }{2} +3x))$

Step-by-step explanation:

Let we consider the given question,

                            $y=sin^2x.cos^3x$

Trigonometric formula for $sin^2 \theta = \frac{(1-cos2\theta)}{2}$ and $cos^3\theta=\frac{(cos3\theta+3cos\theta)}{4}$. Now apply this formula in the above equation.

     $y=sin^2x.cos^3x = \frac{(1-cos2x)}{2}.\frac{(cos3x+3cosx)}{4}$

                               $= \frac{1}{8} (1-cos2x).(cos3x+3cosx)$

                               $= \frac{1}{8} (cos3x+3cosx-cos2xcos3x-3cos2xcosx)$

Now apply $cos A cos B$ formula in the above equation.

               $cos A cos B=\frac{1}{2}[cos(A+B)cos(A-B)]$

So that, the equation is,           $= \frac{1}{8} (cos3x+3cosx-\frac{1}{2} [cos(3x+2x)+cos(3x-2x)]-3*\frac{1}{2} [cos(2x+x)+cos(2x-x)]$

  $= \frac{1}{8} (cos3x+3cosx-\frac{1}{2} [cos(5x)+cos(x)]-3*\frac{1}{2} [cos(3x)+cos(x)]$

  $= \frac{1}{8} \frac{(2cos3x+3*2cosx-cos5x-cosx-3cos3x-3cosx)}{2}$

  $= \frac{1}{16} (2cos3x+6cosx-cos5x-cosx-3cos3x-3cosx)$

$y= \frac{1}{16} (2cosx-cos5x-cos3x)$  ------------------(1)

nth derivative of $cos ax=a^ n .cos(\frac{n\pi }{2} +ax)$

From the above equation,

        $cos x=cos(\frac{n\pi }{2} +x)$  => (a = 1)

       $cos 5x=5^n.cos(\frac{n\pi }{2} +5x)$  => (a = 5)

       $cos 3x=3^n.cos(\frac{n\pi }{2} +3x)$ => (a = 3)

equation(1) can be written as,

Answer: $y= \frac{1}{16} (2cos(\frac{n\pi }{2} +x)-5^n.cos(\frac{n\pi }{2} +5x)-3^n.cos(\frac{n\pi }{2} +3x))$