Find the derivative of cos(4x^2-5)^2

\frac{d}{dx}\left(\cos\left(4x^2-5\right)^2\right)

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Answer

$-16x\cos\left(4x^2-5\right)\sin\left(4x^2-5\right)$

Step by step solution

Problem

$\frac{d}{dx}\left(\cos\left(4x^2-5\right)^2\right)$
1

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

$2\frac{d}{dx}\left(\cos\left(4x^2-5\right)\right)\cos\left(4x^2-5\right)$
2

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 $f'(x) = -\sin(x)\cdot D_x(x)$

$2\left(-1\right)\cos\left(4x^2-5\right)\frac{d}{dx}\left(4x^2-5\right)\sin\left(4x^2-5\right)$
3

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

$2\left(-1\right)\cos\left(4x^2-5\right)\left(\frac{d}{dx}\left(-5\right)+\frac{d}{dx}\left(4x^2\right)\right)\sin\left(4x^2-5\right)$
4

The derivative of the constant function is equal to zero

$2\left(-1\right)\cos\left(4x^2-5\right)\left(0+\frac{d}{dx}\left(4x^2\right)\right)\sin\left(4x^2-5\right)$
5

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$2\left(-1\right)\cos\left(4x^2-5\right)\left(0+4\frac{d}{dx}\left(x^2\right)\right)\sin\left(4x^2-5\right)$
6

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

$2\left(-1\right)\left(0+4\cdot 2x\right)\cos\left(4x^2-5\right)\sin\left(4x^2-5\right)$
7

Multiply $2$ times $4$

$-2\left(0+8x\right)\cos\left(4x^2-5\right)\sin\left(4x^2-5\right)$
8

$x+0=x$, where $x$ is any expression

$-2\cdot 8x\cos\left(4x^2-5\right)\sin\left(4x^2-5\right)$
9

Multiply $8$ times $-2$

$-16x\cos\left(4x^2-5\right)\sin\left(4x^2-5\right)$

Answer

$-16x\cos\left(4x^2-5\right)\sin\left(4x^2-5\right)$

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Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.24 seconds

Views:

98