Nth Derivative Of e2xSin3x.Sin2xe2xSin3x.Sin2x<script type="math/asciimath" id="MathJax-Element-1">e^{2x} Sin3x.Sin2x</script> | Leibnitz Theorem Of Nth Derivatives Examples

To Find n_th derivative of e^(2x) Sin3x Sin2x

In this quick tutorial we will learn how to find n_th derivative of a function which is in the form of e^ax. sinbx. sincx. So let us understand steps to reach n_th derivative very carefully.

Solution: let u = $e^{2 x}$ $e^(2x)$ and v = $Sin 3 x . Sin 2 x$ $Sin3x .Sin2x$

Since we know that:

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

Now let: A = 3X And B = 2X

Therefore,

(A - B)

$(A-B)$ =

X

$X$ and

(A + B)

$(A+B)$ =

5 X

$5X$

Also

y = e^{x} . \frac{1}{2} . {cos (A - B) - cos (A + B)}

$y = e^{x}.\frac{1}{2}.{cos(A-B) - cos(A+B)}$

y = \frac{1}{2} . {e^{x} . cos X - e^{x} . cos 5 X}

$y = \frac{1}{2}.{e^{x}.cosX - e^{x}.cos5X}$

N_{t} h

$N_th$ derivative of

e^{A x} . Cos (B x + c)

$e^{Ax}.Cos(Bx+c)$ is:

{(A^{2} + B^{2})}^{\frac{n}{2}} . e^{A x} . Cos (B x + C + n . {tan}^{- 1} \frac{B}{A})

$(A^{2}+B^{2})^{n/2}.e^{Ax}.Cos(Bx+C+n.tan^{-1}\frac{B}{A})$

n_{t} h

$n_th$ derivative of

\frac{1}{2} . [e^{2 x} . Cos X]

$\frac{1}{2}.[e^{2x}.CosX\]$ is:

\frac{1}{2} . {(2^{2} + 1^{2})}^{\frac{n}{2}} . e^{2 x} . Cos (X + n . {tan}^{- 1} \frac{1}{2})

$\frac{1}{2}.(2^{2}+1^{2})^{n/2}. e^{2x}. Cos(X+n.tan^{-1}\frac{1}{2})$

\frac{1}{2} {.5}^{2} . e^{2 x} . Cos (X + n . {tan}^{- 1} \frac{1}{2})

$\frac{1}{2}.5^{2}. e^{2x}.Cos(X+n.tan^{-1}\frac{1}{2})$

Similarly

n_{t} h

$n_th$ derivative of

e^{2 x} . Cos 5 X

$e^{2x}.Cos5X$ as:

\frac{1}{2} {.29}^{\frac{n}{2}} . e^{2 x} . Cos (5 X + n . {tan}^{- 1} \frac{5}{2})

$\frac{1}{2}.29^{n/2}. e^{2x}.Cos(5X+n.tan^{-1}\frac{5}{2})$

Finally $N_{t} h$ $N_th$ derivative of $e^{2 x} Sin 3 x Sin 2 x$ $e^(2x) Sin3x Sin2x$ is:

\frac{1}{2} {.5}^{2} . e^{2 x} . Cos (X + n . {tan}^{- 1} \frac{1}{2})

$\frac{1}{2}.5^{2}. e^{2x}.Cos(X+n.tan^{-1}\frac{1}{2})$ -

\frac{1}{2} {.29}^{\frac{n}{2}} . e^{2 x} . Cos (5 X + n . {tan}^{- 1} \frac{5}{2})

$\frac{1}{2}.29^{n/2}. e^{2x}.Cos(5X+n.tan^{-1}\frac{5}{2})$

$Nth Derivative Of `e^{2x} Sin3x.Sin2x`$

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Nth Derivative Of $e^{2 x} Sin 3 x . Sin 2 x$ $e^{2x} Sin3x.Sin2x$ | Leibnitz Theorem Of Nth Derivatives Examples

To Find n_th derivative of e^(2x) Sin3x Sin2x

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

Finally $N_{t} h$ $N_th$ derivative of $e^{2 x} Sin 3 x Sin 2 x$ $e^(2x) Sin3x Sin2x$ is:

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Nth Derivative Of e2xSin3x.Sin2xe2xSin3x.Sin2xe^{2x} Sin3x.Sin2x | Leibnitz Theorem Of Nth Derivatives Examples

To Find n_th derivative of e^(2x) Sin3x Sin2x

SinA.SinBSinA.SinBSinA.SinB = [12][12]\[\frac{1}{2}\] {cos(A-B) - cos(A+B)}

Finally NthNthN_th derivative of e2xSin3xSin2xe2xSin3xSin2xe^(2x) Sin3x Sin2x is:

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Nth Derivative Of $e^{2 x} Sin 3 x . Sin 2 x$ $e^{2x} Sin3x.Sin2x$ | Leibnitz Theorem Of Nth Derivatives Examples

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

Finally $N_{t} h$ $N_th$ derivative of $e^{2 x} Sin 3 x Sin 2 x$ $e^(2x) Sin3x Sin2x$ is: