Question

calculus class 12. x^2 tan(x^2-1)​

Answer 1

Answer:

Step-by-step explanation:

Here we are given the function as tanx for which we need to find the Maclaurin’s series. Let us suppose that f(x) = tanx.

We know, Maclaurins series for a function f(x) is given by:

f(x)=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3+⋯ ⋯ .

So let us evaluate the values of f(0), f'(0), f''(0), f'''(0), . . . . . . . . .

For this, we need to evaluate f'(x), f''(x), f'''(x).

We have f(x) = tanx.

Putting x = 0 we get f(0) = tan0.

We know that tan0 is 0 so we get f(0)=0⋯⋯⋯(1) .

Differentiating f(x) with respect to x we get f′(x)=ddxtanx .

We know that, derivative of tanx is sec2x so we have f′(x)=sec2x .

We know that sec2x=1+tan2x so we get f′(x)=1+tan2x .

Putting x = 0 we get f′(0)=1+tan20 .

Using tan0 = 0 we get f′(0)=1+0⇒f′(0)=1⋯⋯⋯(2) .

Differentiating f'(x) with respect to x we get f′′(x)=ddx(1+tan2x) .

We know that d(1)dx=0 so let us find derivative of tan2x . Using chain rule, derivatives of tan2x will be given by 2tan2x⋅ddxtanx=2tanxsec2x .

Therefore we have f′′(x)=0+2tanxsec2x⇒f′′(x)=2tanxsec2x .

Using sec2x=1+tan2x we get f′′(x)=2tanx(1+tan2x) .

Putting x = 0 we get f′′(x0)=2tan0(1+tan20) .

Again using tan0 = 0 we get f′′(0)=2(0)(1+0)=0⇒f′′(0)=0⋯⋯⋯(3) .

Differentiating f''(x) with respect to x we get f′′′(x)=ddx(2tanx(1+tan2x)) .

For this, let us use the product rule. We know that, for any two functions u and v product rule is given as duvdx=uv′+u′v where u' and v' are derivatives of u and v respectively, we get,

f′′′(x)=2(tanxddx(1+tan2x)+ddx(tanx)⋅(1+tan2x))

.

Using previous results for derivatives of (1+tan2x)

and tanx we get,

f′′′(x)=2(tanx(2tanx(1+tan2x))+(1+tan2x)(1+tan2x))

.

Simplifying we get, f′′′(x)=2(2tan2x(1+tan2x)+(1+tan2x)2)

.

Putting x = 0 we get f′′′(0)=2(2tan20(1+tan20)+(1+tan20)2)

.

Again using tan0 = 0 we get,

f′′′(0)=2(2×0(1+0)+(1+0)2)⇒f′′′(0)=2(0+12)⇒f′′′(0)=2(1)⇒f′′′(0)=2⋯⋯⋯(4)

Putting all these values in the Maclaurins series we get,

tanx=0+(1)x+02!x2+23!x3+⋯ ⋯ .

Simplifying we get tanx=x+23!x3+⋯ ⋯ .

Expanding 3! as 3×2 we get tanx=x+23×2x3+⋯ ⋯⇒tanx=x+x33+⋯ ⋯ .

Therefore, the Maclaurins series for tanx is given as tanx=x+x33