Derivatives of trigonometric functions Calculator

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Example

Solved Problems

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Solved example of derivatives of trigonometric functions

$\frac{d}{dx}\cos\left(3x^2+x-5\right)$

The derivative of the cosine of a function is equal to minus the sine of the function times the derivative of the function, in other words, if $f(x) = \cos(x)$, then $f'(x) = -\sin(x)\cdot D_x(x)$

$-\sin\left(3x^2+x-5\right)\frac{d}{dx}\left(3x^2+x-5\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$-\sin\left(3x^2+x-5\right)\left(\frac{d}{dx}\left(3x^2\right)+\frac{d}{dx}\left(x-5\right)\right)$

The derivative of a function multiplied by a constant ($3$) is equal to the constant times the derivative of the function

$-\sin\left(3x^2+x-5\right)\left(3\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(x-5\right)\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$-\sin\left(3x^2+x-5\right)\left(6x+\frac{d}{dx}\left(x-5\right)\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$-\sin\left(3x^2+x-5\right)\left(6x+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(-5\right)\right)$

The derivative of the constant function ($-5$) is equal to zero

$-\sin\left(3x^2+x-5\right)\left(6x+\frac{d}{dx}\left(x\right)\right)$

The derivative of the linear function is equal to $1$

$-\sin\left(3x^2+x-5\right)\left(6x+1\right)$

Final Answer