Question Statement Differentiate the following with respect to the variables involved:
x 2 sec β‘ ( 4 x ) x^{2} \sec(4x) x 2 sec ( 4 x ) tan β‘ 2 ( ΞΈ ) sec β‘ 2 ( ΞΈ ) \tan^2(\theta) \sec^2(\theta) tan 2 ( ΞΈ ) sec 2 ( ΞΈ ) ( sin β‘ 2 ΞΈ β cos β‘ 3 ΞΈ ) 2 (\sin 2\theta - \cos 3\theta)^{2} ( sin 2 ΞΈ β cos 3 ΞΈ ) 2 cos β‘ ( x ) + sin β‘ ( x ) \cos(\sqrt{x}) + \sqrt{\sin(x)} cos ( x β ) + sin ( x ) β
Background and Explanation To differentiate these functions, youβll need to apply rules such as the product rule, chain rule, and standard derivatives of trigonometric functions. Make sure to keep in mind how each function interacts with its variable, especially with composite functions.
Solution i. x 2 sec β‘ ( 4 x ) x^{2} \sec(4x) x 2 sec ( 4 x )
Let y = x 2 sec β‘ ( 4 x ) y = x^{2} \sec(4x) y = x 2 sec ( 4 x )
To differentiate this, we apply the product rule :
d d x [ x 2 sec β‘ ( 4 x ) ] = x 2 d d x [ sec β‘ ( 4 x ) ] + sec β‘ ( 4 x ) d d x [ x 2 ] \frac{d}{dx}[x^{2} \sec(4x)] = x^{2} \frac{d}{dx}[\sec(4x)] + \sec(4x) \frac{d}{dx}[x^{2}] d x d β [ x 2 sec ( 4 x )] = x 2 d x d β [ sec ( 4 x )] + sec ( 4 x ) d x d β [ x 2 ]
Differentiate sec β‘ ( 4 x ) \sec(4x) sec ( 4 x )
d d x [ sec β‘ ( 4 x ) ] = sec β‘ ( 4 x ) tan β‘ ( 4 x ) Γ 4 \frac{d}{dx}[\sec(4x)] = \sec(4x) \tan(4x) \times 4 d x d β [ sec ( 4 x )] = sec ( 4 x ) tan ( 4 x ) Γ 4 d d x [ x 2 ] = 2 x \frac{d}{dx}[x^{2}] = 2x d x d β [ x 2 ] = 2 x Substituting these into the equation:
d d x [ x 2 sec β‘ ( 4 x ) ] = x 2 β
4 sec β‘ ( 4 x ) tan β‘ ( 4 x ) + sec β‘ ( 4 x ) β
2 x \frac{d}{dx}[x^{2} \sec(4x)] = x^{2} \cdot 4 \sec(4x) \tan(4x) + \sec(4x) \cdot 2x d x d β [ x 2 sec ( 4 x )] = x 2 β
4 sec ( 4 x ) tan ( 4 x ) + sec ( 4 x ) β
2 x Thus, the derivative is:
2 x sec β‘ ( 4 x ) [ 2 x tan β‘ ( 4 x ) + 1 ] \boxed{2x \sec(4x) [2x \tan(4x) + 1]} 2 x sec ( 4 x ) [ 2 x tan ( 4 x ) + 1 ] β ii. tan β‘ 2 ( ΞΈ ) sec β‘ 2 ( \thetae ) \tan^2(\theta) \sec^2(\thetae) tan 2 ( ΞΈ ) sec 2 ( \thetae )
Differentiate with respect to ΞΈ \theta ΞΈ
We apply the product rule :
d d β
[ tan β‘ 2 ( β
) sec β‘ 2 ( β
) ] = sec β‘ 2 ( β
) d d β
[ tan β‘ 2 ( β
) ] + tan β‘ 2 ( β
) d d β
[ sec β‘ 2 ( β
) ] \frac{d}{d\emptyset}[\tan^2(\emptyset) \sec^2(\emptyset)] = \sec^2(\emptyset) \frac{d}{d\emptyset}[\tan^2(\emptyset)] + \tan^2(\emptyset) \frac{d}{d\emptyset}[\sec^2(\emptyset)] d β
d β [ tan 2 ( β
) sec 2 ( β
)] = sec 2 ( β
) d β
d β [ tan 2 ( β
)] + tan 2 ( β
) d β
d β [ sec 2 ( β
)]
The derivative of tan β‘ 2 ( ΞΈ ) \tan^2(\theta) tan 2 ( ΞΈ )
d d β
[ tan β‘ 2 ( β
) ] = 2 tan β‘ ( β
) sec β‘ 2 ( β
) \frac{d}{d\emptyset}[\tan^2(\emptyset)] = 2 \tan(\emptyset) \sec^2(\emptyset) d β
d β [ tan 2 ( β
)] = 2 tan ( β
) sec 2 ( β
)
The derivative of sec β‘ 2 ( ΞΈ ) \sec^2(\theta) sec 2 ( ΞΈ )
d d β
[ sec β‘ 2 ( β
) ] = 2 sec β‘ ( β
) sec β‘ ( β
) tan β‘ ( β
) \frac{d}{d\emptyset}[\sec^2(\emptyset)] = 2 \sec(\emptyset) \sec(\emptyset) \tan(\emptyset) d β
d β [ sec 2 ( β
)] = 2 sec ( β
) sec ( β
) tan ( β
) Now substitute into the product rule:
d d β
[ tan β‘ 2 ( β
) sec β‘ 2 ( β
) ] = sec β‘ 2 ( β
) [ 2 tan β‘ ( β
) sec β‘ 2 ( β
) ] + tan β‘ 2 ( β
) [ 2 sec β‘ 2 ( β
) tan β‘ ( β
) ] \frac{d}{d\emptyset}[\tan^2(\emptyset) \sec^2(\emptyset)] = \sec^2(\emptyset) [2 \tan(\emptyset) \sec^2(\emptyset)] + \tan^2(\emptyset) [2 \sec^2(\emptyset) \tan(\emptyset)] d β
d β [ tan 2 ( β
) sec 2 ( β
)] = sec 2 ( β
) [ 2 tan ( β
) sec 2 ( β
)] + tan 2 ( β
) [ 2 sec 2 ( β
) tan ( β
)] Simplifying:
3 tan β‘ 2 ( β
) sec β‘ 4 ( β
) + 2 sec β‘ 2 ( β
) tan β‘ 4 ( β
) \boxed{3 \tan^2(\emptyset) \sec^4(\emptyset) + 2 \sec^2(\emptyset) \tan^4(\emptyset)} 3 tan 2 ( β
) sec 4 ( β
) + 2 sec 2 ( β
) tan 4 ( β
) β iii. ( sin β‘ 2 ΞΈ β cos β‘ 3 ΞΈ ) 2 (\sin 2\theta - \cos 3\theta)^{2} ( sin 2 ΞΈ β cos 3 ΞΈ ) 2
Let y = ( sin β‘ 2 ΞΈ β cos β‘ 3 ΞΈ ) 2 y = (\sin 2\theta - \cos 3\theta)^2 y = ( sin 2 ΞΈ β cos 3 ΞΈ ) 2
To differentiate, we apply the chain rule :
d d β
[ ( sin β‘ 2 β
β cos β‘ 3 β
) 2 ] = 2 ( sin β‘ 2 β
β cos β‘ 3 β
) d d β
[ sin β‘ 2 β
β cos β‘ 3 β
] \frac{d}{d\emptyset} \left[(\sin 2\emptyset - \cos 3\emptyset)^2\right] = 2 (\sin 2\emptyset - \cos 3\emptyset) \frac{d}{d\emptyset} [\sin 2\emptyset - \cos 3\emptyset] d β
d β [ ( sin 2β
β cos 3β
) 2 ] = 2 ( sin 2β
β cos 3β
) d β
d β [ sin 2β
β cos 3β
] Now differentiate the inner expression sin β‘ 2 β
β cos β‘ 3 β
\sin 2\emptyset - \cos 3\emptyset sin 2β
β cos 3β
d d β
[ sin β‘ 2 β
β cos β‘ 3 β
] = 2 cos β‘ 2 β
+ 3 sin β‘ 3 β
\frac{d}{d\emptyset} [\sin 2\emptyset - \cos 3\emptyset] = 2 \cos 2\emptyset + 3 \sin 3\emptyset d β
d β [ sin 2β
β cos 3β
] = 2 cos 2β
+ 3 sin 3β
Substituting everything back:
2 ( sin β‘ 2 β
β cos β‘ 3 β
) [ 2 cos β‘ 2 β
+ 3 sin β‘ 3 β
] \boxed{2 (\sin 2\emptyset - \cos 3\emptyset) [2 \cos 2\emptyset + 3 \sin 3\emptyset]} 2 ( sin 2β
β cos 3β
) [ 2 cos 2β
+ 3 sin 3β
] β iv. cos β‘ ( x ) + sin β‘ ( x ) \cos(\sqrt{x}) + \sqrt{\sin(x)} cos ( x β ) + sin ( x ) β
Let y = cos β‘ ( x ) + sin β‘ ( x ) y = \cos(\sqrt{x}) + \sqrt{\sin(x)} y = cos ( x β ) + sin ( x ) β
Differentiate each term:
For cos β‘ ( x ) \cos(\sqrt{x}) cos ( x β ) chain rule :
d d x [ cos β‘ ( x ) ] = β sin β‘ ( x ) β
1 2 x \frac{d}{dx}[\cos(\sqrt{x})] = -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} d x d β [ cos ( x β )] = β sin ( x β ) β
2 x β 1 β
For sin β‘ ( x ) \sqrt{\sin(x)} sin ( x ) β
d d x [ sin β‘ ( x ) ] = 1 2 sin β‘ ( x ) β
cos β‘ ( x ) \frac{d}{dx}[\sqrt{\sin(x)}] = \frac{1}{2\sqrt{\sin(x)}} \cdot \cos(x) d x d β [ sin ( x ) β ] = 2 sin ( x ) β 1 β β
cos ( x ) Now, combine the results:
d d x [ cos β‘ ( x ) + sin β‘ ( x ) ] = β sin β‘ ( x ) 2 x + cos β‘ ( x ) 2 sin β‘ ( x ) \frac{d}{dx}[\cos(\sqrt{x}) + \sqrt{\sin(x)}] = -\frac{\sin(\sqrt{x})}{2\sqrt{x}} + \frac{\cos(x)}{2\sqrt{\sin(x)}} d x d β [ cos ( x β ) + sin ( x ) β ] = β 2 x β sin ( x β ) β + 2 sin ( x ) β cos ( x ) β Thus, the derivative is:
1 2 ( sin β‘ ( x ) x + cos β‘ ( x ) sin β‘ ( x ) ) \boxed{\frac{1}{2} \left(\frac{\sin(\sqrt{x})}{\sqrt{x}} + \frac{\cos(x)}{\sqrt{\sin(x)}}\right)} 2 1 β ( x β sin ( x β ) β + sin ( x ) β cos ( x ) β ) β
Product Rule : d d x [ u ( x ) v ( x ) ] = u β² ( x ) v ( x ) + u ( x ) v β² ( x ) \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) d x d β [ u ( x ) v ( x )] = u β² ( x ) v ( x ) + u ( x ) v β² ( x ) Chain Rule : d d x [ f ( g ( x ) ) ] = f β² ( g ( x ) ) β
g β² ( x ) \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) d x d β [ f ( g ( x ))] = f β² ( g ( x )) β
g β² ( x ) Standard derivatives of trigonometric functions:
d d x [ sec β‘ x ] = sec β‘ x tan β‘ x \frac{d}{dx}[\sec x] = \sec x \tan x d x d β [ sec x ] = sec x tan x d d x [ tan β‘ x ] = sec β‘ 2 x \frac{d}{dx}[\tan x] = \sec^2 x d x d β [ tan x ] = sec 2 x d d x [ sin β‘ x ] = cos β‘ x \frac{d}{dx}[\sin x] = \cos x d x d β [ sin x ] = cos x d d x [ cos β‘ x ] = β sin β‘ x \frac{d}{dx}[\cos x] = -\sin x d x d β [ cos x ] = β sin x
Summary of Steps
For x 2 sec β‘ ( 4 x ) x^{2} \sec(4x) x 2 sec ( 4 x ) :
Apply product rule and chain rule.
Differentiate each part and combine the results.
For tan β‘ 2 ( β
) sec β‘ 2 ( β
) \tan^2(\emptyset) \sec^2(\emptyset) tan 2 ( β
) sec 2 ( β
) :
Apply product rule.
Differentiate tan β‘ 2 ( β
) \tan^2(\emptyset) tan 2 ( β
) sec β‘ 2 ( β
) \sec^2(\emptyset) sec 2 ( β
)
For ( sin β‘ 2 β
β cos β‘ 3 β
) 2 (\sin 2\emptyset - \cos 3\emptyset)^{2} ( sin 2β
β cos 3β
) 2 :
Use chain rule.
Differentiate the inner expression and apply the result to the outer square.
For cos β‘ ( x ) + sin β‘ ( x ) \cos(\sqrt{x}) + \sqrt{\sin(x)} cos ( x β ) + sin ( x ) β :
Use chain rule to differentiate each term.
Combine the results for the final derivative.
