Question

Find the derivative of the following

$sin( {x}^{2} + 5)$
$cos \sqrt{x}$
$cos(sin \: x)$

Brief explaination needed.

Hoping for great answers ​

Answer 1

$\huge\boxed{\mathtt\red{SOLUTION\::-}}$

PROVIDED :-

1. $\large\mathtt{sin( {x}^{2} + 5)}$
2. $\large\mathtt{cos \sqrt{x}}$
3. $\large\mathtt{cos(sin \: x)}$

TO FIND :-

• derivatives of the given problems

SOLVING :-

$\large\boxed{\mathtt\purple{Answer\:1\::-}}$

✒ $\large{\mathtt{\frac{d}{dx}(sin(x²+5))}}$

Applying the chain rule :-

→ y = sin u

→ u = x²+5

✒ $\large{\mathtt{\frac{dy}{dx}=\frac{dy}{du}×\frac{du}{dx}}}$

✒ $\large{\mathtt{\frac{dy}{dx}=\frac{sin\:u}{du}×\frac{d(x²+5)}{dx}}}$

✒ $\large{\mathtt{\frac{dy}{dx}= cos\:u×(2x+0)}}$

✒ $\large{\mathtt{\frac{dy}{dx}=cos(x²+5)×2x}}$

$\large\boxed{\mathtt\green{∴derivation\: of \:sin(x²+5) \:is\: 2x•cos(x²+5)}}$

$\large\boxed{\mathtt\purple{Answer\:2\::-}}$

✒ $\large{\mathtt{\frac{d}{dx}(cos(√x))}}$

Applying chain rule :-

→ y = cos u

→ u = √x

✒ $\large{\mathtt{\frac{dy}{dx}=\frac{dy}{du}×\frac{du}{dx}}}$

✒ $\large{\mathtt{\frac{dy}{dx}=\frac{d(cos\:u)}{du}×\frac{d(√x)}{dx}}}$

✒ $\large{\mathtt{\frac{dy}{dx}= -\:sin(√x)×\frac{1}{2√x}}}$

$\large\boxed{\mathtt\green{∴the\: derivation\:of\:cos(√x)\:is\:\frac{- sin(√x)}{2√x}}}$

$\large\boxed{\mathtt\purple{Answer\:3\::-}}$

✒ $\large{\mathtt{\frac{d}{dx}(cos(sin\:x)))}}$

Applying the chain rule :-

→ y = cos u

→ u = sin x

✒ $\large{\mathtt{\frac{dy}{dx}=\frac{dy}{du}×\frac{du}{dx}}}$

✒ $\large{\mathtt{\frac{dy}{dx}=\frac{cos(u)}{du}×\frac{d(sin\:x)}{dx}}}$

✒ $\large{\mathtt{\frac{dy}{dx}=-sin(u)×cos(x)}}$

✒ $\large{\mathtt{\frac{dy}{dx}=-sin(sin\:x)×cos(x)}}$

$\large\boxed{\mathtt\green{∴the\: derivation\:of\:cos(sin\:x)\:is\:-sin(sin\:x)•cos(x)}}$

$\huge\boxed{\mathtt\green{CONCLUSION\::-}}$

Derivatives of the above problems are :-

1. $\large{\mathtt{2x•cos(x²+5)}}$
2. $\large{\mathtt{\bold{\frac{-sin(√x)}{2√x}}}}$
3. $\large{\mathtt{-sin(sin\:x)•cos(x)}}$