Question

please it's very urgent​

Answer 1

Answer:

$\sf f'(x)=1$

Step-by-step explanation:

Given:

$\sf f(x)=\bigg(sin^{-1} \bigg(\dfrac{3\:cosx+4\:sinx}{5} \bigg)\bigg)$

To Find:

$\sf f'(x)$

Solution:

Here we have to differentiate the function.

First simplifying the function,

$\sf sin^{-1} \bigg( \dfrac{3}{5}\:cosx+\dfrac{4}{5}\:sinx\bigg)$

Let us assume 3/5 = sin θ------(1)

We know,

sin θ = opposite/hypotenuse

Considering a right angled triangle to find the adjacent side,

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf 3}\put(4,3){\large \bf 5}}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf \Theta }\end{picture}$

By Pythagoras theorem,

$\sf Adjacent\:side=\sqrt{Hypotenuse^{2}- Opposite^{2}}$

Therefore,

Adjacent side = √(25 - 9) = 4

Now we know that,

cos θ = adjacent/hypotenuse = 4/5-----(2)

Substituting 2 and 1 in the above equation,

$\sf sin^{-1} \bigg( sin\:\theta\:cosx+cos\:\theta\:sinx\bigg)$

Now we know that,

$\boxed{\sf sinx\:cos\:y+cos\:x\:sin\:y=sin(x+y)}$

Applying the identity,

$\sf sin^{-1} \bigg( sin\:(\theta+x)\bigg)$

Cancelling the trignometric functions,

$\implies \theta+x$

Now if sin θ = 3/5,

θ = sin ⁻¹ 3/5

Substitute the value,

$\sf \implies sin^{-1} \bigg(\dfrac{3}{5} \bigg)+x$

Now differentiating it with respect to x,

We know that,

$\boxed{\sf \dfrac{d}{dx} (constant)=0}$

Also,

$\boxed{\sf \dfrac{d}{dx} (x)=1}$

Applying the identities,

$\sf f'(x)=0+1$

$\implies =1$

Hence the derivative of the given function is 1.

Answer 2

Answer:

\sf f'(x)=1f′(x)=1

Step-by-step explanation:

Given:

\sf f(x)=\bigg(sin^{-1} \bigg(\dfrac{3\:cosx+4\:sinx}{5} \bigg)\bigg)f(x)=(sin−1(53cosx+4sinx))

To Find:

\sf f'(x)f′(x)

Solution:

Here we have to differentiate the function.

First simplifying the function,

\sf sin^{-1} \bigg( \dfrac{3}{5}\:cosx+\dfrac{4}{5}\:sinx\bigg)sin−1(53cosx+54sinx)

Let us assume 3/5 = sin θ------(1)

We know,

sin θ = opposite/hypotenuse

Considering a right angled triangle to find the adjacent side,

\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf 3}\put(4,3){\large \bf 5}}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf $\Theta$}\end{picture}

By Pythagoras theorem,

\sf Adjacent\:side=\sqrt{Hypotenuse^{2}- Opposite^{2}}Adjacentside=Hypotenuse2−Opposite2

Therefore,

Adjacent side = √(25 - 9) = 4

Now we know that,

cos θ = adjacent/hypotenuse = 4/5-----(2)

Substituting 2 and 1 in the above equation,

\sf sin^{-1} \bigg( sin\:\theta\:cosx+cos\:\theta\:sinx\bigg)sin−1(sinθcosx+cosθsinx)

Now we know that,

\boxed{\sf sinx\:cos\:y+cos\:x\:sin\:y=sin(x+y)}sinxcosy+cosxsiny=sin(x+y)

Applying the identity,

\sf sin^{-1} \bigg( sin\:(\theta+x)\bigg)sin−1(sin(θ+x))

Cancelling the trignometric functions,

\implies \theta+x⟹θ+x

Now if sin θ = 3/5,

θ = sin ⁻¹ 3/5

Substitute the value,

\sf \implies sin^{-1} \bigg(\dfrac{3}{5} \bigg)+x⟹sin−1(53)+x

Now differentiating it with respect to x,

We know that,

\boxed{\sf \dfrac{d}{dx} (constant)=0}dxd(constant)=0

Also,

\boxed{\sf \dfrac{d}{dx} (x)=1}dxd(x)=1

Applying the identities,

\sf f'(x)=0+1f′(x)=0+1

\implies =1⟹=1

Hence the derivative of the given function is 1.