Question

Find the nth derivative of sin5x cos4x​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{y=sin(5x)\,cos(4x)}$

$\tt{\implies\,y=\dfrac{1}{2}\cdot2\,sin(5x)\,cos(4x)}$

$\tt{\implies\,y=\dfrac{1}{2}\left\{sin\left(5x+4x\right)+sin\left(5x-4x\right)\right\}}$

$\tt{\implies\,y=\dfrac{1}{2}\left\{sin\left(9x\right)+sin\left(x\right)\right\}}$

$\tt{\implies\,y=\dfrac{1}{2}\,sin(9x)+\dfrac{1}{2}\,sin(x)}$

We know,

$\boxed{\bf{\dfrac{{d}^{n}}{d{x}^{n}}\Big(sin(ax+b)\Big)={a}^{n}\cdot\,sin\left(ax+b+\dfrac{n\pi}{2}\right)}}$

So,

$\tt{\implies\,y_{n}=\dfrac{1}{2}\cdot{9}^{n}\,sin\left(9x+\dfrac{n\pi}{2}\right)+\dfrac{1}{2}\,sin\left(x+\dfrac{n\pi}{2}\right)}$

$\tt{\implies\,y_{n}=\dfrac{1}{2}\left\{{9}^{n}\cdot\,sin\left(9x+\dfrac{n\pi}{2}\right)+sin\left(x+\dfrac{n\pi}{2}\right)\right\}}$

Answer 2

I =1/3cos3x−cosx−2cosx+c is the derivative of sin5xcos4x.

Given: sin5xcos4x
To find: nth derivative
Solution:

• Derivatives are delineated as the variable rate of change of a function concerning an free changing.
• The derivative is generally secondhand when skilled is few variable capacity, and the rate of change is not fixed. The derivative is used to measure the subtlety of individual changing (reliant changing) concerning another changeable (free changeable).
• In this item, we are make use of explain what are derivatives, the description of products Math, limits and products painstakingly.
• Derivatives in Maths refers to the immediate rate of change of a quota concerning the added. It helps to scrutinize the importance by importance character of any.
  
  Lets simplify the equation:
  ∫sin5xcos4xdx

sin5x=sin4xsinx=(sin2x)2sinx=(1−cos2x)2sinx

Let t=cosx

dt=−sinxdx

−dt=sinxdx

Now,

I=∫(1−cos2x)2sinxcos4xdx

I=−∫(1−t2)2t4dt

I=−∫1+t2−2tt4dt

I=−∫(t−4+t−2−2t−3)dt

I=−[t−3−3+t+2t−1]+c

I=−[(cosx)−3−3+cosx+2(cosx)−1]+c

I=1/3cos3x−cosx−2cosx+c

So we get our derivative as 1/3cos3x−cosx−2cosx+c

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