Find the derivative of the following functions.
f:x\longmapsto 3\cos\left(x\right)
The derivative of a cosine function is:
\dfrac{d}{dx}\cos\left(f\left(x\right)\right) = -f'\left(x\right)\sin\left(f\left(x\right)\right)
Therefore:
\dfrac{d}{dx}\left[3\cos\left(x\right)\right]= 3\dfrac{d}{dx}\left[\cos\left(x\right)\right]=3\left(-\dfrac{d}{dx}x\right)\sin\left(x\right)=-3\sin\left(x\right)
For any x \in \mathbb{R} , f is differentiable and f'\left(x\right) = -3\sin\left(x\right)
f : x \longmapsto \sin\left(3x + 2\right)
The derivative of a sine function is:
\dfrac{d}{dx}\sin\left(f\left(x\right)\right) = f'\left(x\right)\cos\left(f\left(x\right)\right)
Therefore:
\dfrac{d}{dx}\left[ \sin\left(3x + 2\right) \right]=\left[ \dfrac{d}{dx}\left(3x + 2\right) \right]\cos\left(3x + 2\right)=3\cos\left(3x + 2\right)
For any x \in \mathbb{R} , f is differentiable and f'\left(x\right) = 3\cos\left(3x + 2\right)
f : x \longmapsto \cos\left(\sin\left(x\right)\right)
The derivatives of a sine and cosine function are, respectively:
\dfrac{d}{dx}\sin\left(f\left(x\right)\right) = f'\left(x\right)\cos\left(f\left(x\right)\right)
\dfrac{d}{dx}\cos\left(f\left(x\right)\right) = -f'\left(x\right)\sin\left(f\left(x\right)\right)
Therefore:
\dfrac{d}{dx}\left[ \cos\left(\sin\left(x\right)\right) \right]=-\left[\dfrac{d}{dx}\sin\left(x\right) \right]\sin\left(\sin\left(x\right)\right)=-\cos\left(x\right)\sin\left(\sin\left(x\right)\right)
For any x \in \mathbb{R} , f is differentiable and f'\left(x\right) = -\cos\left(x\right)\sin\left(\sin\left(x\right)\right)
x \longmapsto -6\tan\left(\dfrac{1}{3}x\right)
The tangent can be expressed as the quotient of sine and cosine:
\dfrac{d}{dx}\tan\left(f\left(x\right)\right) = \dfrac{d}{dx}\dfrac{\sin\left(f\left(x\right)\right)}{\cos\left(f\left(x\right)\right)} for any real x such that f\left(x\right) \neq \dfrac{\pi}{2}+n\pi (where n is a non-zero integer)
Then, with the formula for the derivative of a quotient:
\dfrac{d}{dx}\tan\left(f\left(x\right)\right) = \dfrac{\dfrac{d}{dx}\left[\sin\left(f\left(x\right)\right)\right] \cos\left(f\left(x\right)\right) - \sin\left(f\left(x\right)\right) \dfrac{d}{dx}\left[\cos\left(f\left(x\right)\right)\right]}{\cos\left(f\left(x\right)\right)^2}
Finally:
\dfrac{d}{dx}\tan\left(f\left(x\right)\right) = \dfrac{f'\left(x\right) \left(\cos^2\left(f\left(x\right)\right) + \sin^2\left(f\left(x\right)\right)\right)}{\cos\left(f\left(x\right)\right)^2} = \dfrac{f'\left(x\right)}{\cos\left(f\left(x\right)\right)^2}
The derivative of a tangent function is then:
\dfrac{d}{dx}\tan\left(f\left(x\right)\right) = f'\left(x\right)\sec^{2}\left(f\left(x\right)\right)
Where:
\sec\left(x\right) = \dfrac{1}{\cos\left(x\right)}.
Therefore for any real x such that \dfrac{x}{3}\neq \dfrac{\pi}{2}+n\pi i.e. x\neq\left(\dfrac{3}{2}+3n\right)\pi (where n is a non-zero integer):
\dfrac{d}{dx}\left[ -6\tan\left(\dfrac{1}{3}x\right) \right]=-6\left[ \dfrac{d}{dx}\left(\dfrac{1}{3}x\right) \right]\sec^{2}\left(\dfrac{1}{3}x\right)=-6\left(\dfrac{1}{3}\right)\sec^{2}\left(\dfrac{1}{3}x\right)=-2\sec^{2}\left(\dfrac{1}{3}x\right)
For any real x such that x\neq\left(\dfrac{3}{2}+3n\right)\pi (where n is a non-zero integer), f is differentiable and f'\left(x\right) = -2\sec^{2}\left(\dfrac{1}{3}x\right)
f : x \longmapsto \cos^2\left(3x\right)
The derivatives of a sin and cosine function are:
\dfrac{d}{dx}\sin\left(f\left(x\right)\right) = f'\left(x\right)\cos\left(f\left(x\right)\right)
\dfrac{d}{dx}\cos\left(f\left(x\right)\right) = -f'\left(x\right)\sin\left(f\left(x\right)\right)
Also:
\dfrac{d}{dx}f^2\left(x\right) = 2f'\left(x\right)f\left(x\right)
Therefore:
\dfrac{d}{dx}\left[ \cos^2\left(3x\right) \right]
=2\left[ \dfrac{d}{dx}\cos\left(3x\right) \right] \left[ \cos\left(3x\right) \right]
=2\left[ \right[\dfrac{d}{dx}\left(3x\right)\left] \left(-\sin\left(3x\right)\right) \right] \left[ \cos\left(3x\right) \right]
=-2\left[ 3\sin\left(3x\right) \right] \left[ \cos\left(3x\right) \right]=-6\sin\left(3x\right)\cos\left(3x\right)
For any x \in \mathbb{R} , f is differentiable and f'\left(x\right) = -6\sin\left(3x\right)\cos\left(3x\right)
f : x \longmapsto \dfrac{1}{5}\tan\left(-4x\right)
The tangent can be expressed as the quotient of sine and cosine:
\dfrac{d}{dx}\tan\left(f\left(x\right)\right) = \dfrac{d}{dx}\dfrac{\sin\left(f\left(x\right)\right)}{\cos\left(f\left(x\right)\right)} for any real x such that f\left(x\right) \neq \dfrac{\pi}{2}+n\pi (where n is a non-zero integer)
Then, with the formula for the derivative of a quotient:
\dfrac{d}{dx}\tan\left(f\left(x\right)\right) = \dfrac{\dfrac{d}{dx}\left[\sin\left(f\left(x\right)\right)\right] \cos\left(f\left(x\right)\right) - \sin\left(f\left(x\right)\right) \dfrac{d}{dx}\left[\cos\left(f\left(x\right)\right)\right]}{\cos\left(f\left(x\right)\right)^2}
Finally:
\dfrac{d}{dx}\tan\left(f\left(x\right)\right) = \dfrac{f'\left(x\right) \left(\cos^2\left(f\left(x\right)\right) + \sin^2\left(f\left(x\right)\right)\right)}{\cos\left(f\left(x\right)\right)^2} = \dfrac{f'\left(x\right)}{\cos\left(f\left(x\right)\right)^2}
The derivative of a tangent function is then:
\dfrac{d}{dx}\tan\left(f\left(x\right)\right) = \dfrac{f'\left(x\right)}{\cos\left(f\left(x\right)\right)^2} = f'\left(x\right)\sec^{2}\left(f\left(x\right)\right)
Where:
\sec\left(x\right) = \dfrac{1}{\cos\left(x\right)}
Therefore, for any real x such that -4x \neq \dfrac{\pi}{2}+n\pi i.e. x\neq -\left(\dfrac{1}{8}+\dfrac{n}{4}\right)\pi (where n is a non-zero integer):
\dfrac{d}{dx}\left[ \dfrac{1}{5}\tan\left(-4x\right) \right]=\dfrac{1}{5}\left[ \dfrac{d}{dx}\left(-4x\right) \right]\sec^{2}\left(-4x\right)
\dfrac{d}{dx}\left[ \dfrac{1}{5}\tan\left(-4x\right) \right]=\dfrac{1}{5}\left(-4\right)\sec^{2}\left(-4x\right)
\dfrac{d}{dx}\left[ \dfrac{1}{5}\tan\left(-4x\right) \right]=\dfrac{-4}{5}\sec^{2}\left(-4x\right)
And since \sec^{2}\left(a\right) = \sec^{2}\left(-a\right), where a is a real number:
\dfrac{-4}{5}\sec^{2}\left(-4x\right) = \dfrac{-4}{5}\sec^{2}\left(4x\right)
For any real x such that x\neq -\left(\dfrac{1}{8}+\dfrac{n}{4}\right)\pi (where n is a non-zero integer), f is differentiable and f'\left(x\right) = \dfrac{-4}{5}\sec^{2}\left(4x\right)
f : x \longmapsto \dfrac{1}{2}\sin\left(6x^2\right)
The derivative of a sine function is:
\dfrac{d}{dx}\sin\left(f\left(x\right)\right) = f'\left(x\right)\cos\left(f\left(x\right)\right)
Therefore:
\dfrac{d}{dx}\left[ \dfrac{1}{2}\sin\left(6x^2\right) \right]
=\dfrac{1}{2}\left[ \dfrac{d}{dx}\left(6x^2\right) \right]\cos\left(6x^2\right)
=\dfrac{1}{2}\left(12x\right)\cos\left(6x^2\right)
=6x\cos\left(6x^2\right)
For any x \in \mathbb{R} , f is differentiable and f'\left(x\right) = 6x\cos\left(6x^2\right)