In calculus, finding the derivative of a function means figuring out how the function changes at any point. This process is called differentiation. The derivative tells us the rate of change or the slope of the function at a given value of x.

Let’s take the function sinx·cosx. Its derivative is cos(2x). This result can be found using different rules in calculus.

One way to find this is by using the product rule, because we are multiplying two functions: sinx and cosx. Another way is to use the first principle of derivatives, which is the basic definition of a derivative using limits.

Apart from this, we can also find derivatives of more complex functions like:

• (sin⁡x)cos⁡x : This is a combination of a power function, and we use logarithmic differentiation to solve it.
• sinx + cosx: For this, we use basic derivative rules and simply find the derivative of each part separately.

What is Derivative of Sinx Cosx?

The derivative of sinx cosx gives us the rate of change of the function of the product of sinx cosx with respect to the variable x.

Derivative of sinx cosx is given by \(\frac{d}{dx}(\sin x\cos x)=\cos 2x\)

We can calculate the derivative of sinx cosx by 2 methods:

1. By First Principle
2. By Product Rule

First principle: It is also known as the delta method and refers to the general expression for the slope of a curve

\(f^{\prime}(x)=\frac{dy}{dx}=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\)

Product rule: It is the method to find the derivative of the product of the two functions.

Given the formula,

\(\frac{d}{dx}[f(x).g(x)]=f^{\prime}(x).g(x)+f(x).g^{\prime}(x)\)

The derivative of sin(x) cos(x) represented as d/dx [sin(x) cos(x)] or as (sin(x)cos(x))' is given by the formula: 

d/dx [sin(x) cos(x)] =(sin(x)cos(x))'= \(cos^2(x)-sin^2(x)\)= cos(2x)

We can find the derivative of the function sin(x) cos(x) at any point x=a easily with the help of the above formula by simply substituting the value of x=a in cos(2x).

Now let's see how this formula is derived using first principle and product rule.

Derivative of Sinx Cosx by First Principle

We will be proving the result \(\frac{d}{dx}(\sin x\cos x) =\cos 2x\) by the first principle, taking the values of our function at x and x+h to the limiting values as h approaches to 0.

By first principle we have the formula

\(f^{\prime}(x)=\frac{dy}{dx}=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\)

\((\sin x\ \cos x)^{\prime}=\frac{lim_{h\rightarrow 0}[\sin (x+h)\cos (x+h)-\sin x\cos x]}{h}\)- (1)

Referring to formulae,

\(\sin (A+B)=\sin A\cos B+\cos A\sin B\) and

\(\cos (A+B)=\cos A\cos B-\sin A\sin B\)

We get equation (1) as,

\(lim_{h\rightarrow0}=\frac{[(\sin x\cosh+\sinh\cos x)(\cos x\cosh-\sin x\sinh)-\sin x\cos x]}{h}\)

\(lim_{h→0}\frac{[sinx cos x cos2h + sinh cosh cos2x – sin2x sinh cosh – cosx sinx sin2h – sinx cosx]}{h}\)

\(\frac{lim_{h→0}[(sinx cos x cos2h – cosx sinx sin2h – sinx cosx) + sinh cosh cos2x – sin2x sinh cosh]}{h}\)

\(\frac{lim_{h→0}[(sinx cos x cos2h – cosx sinx sin2h – sinx cosx) + sinh cosh cos2x – sin2x sinh cosh]}{h}\)

\( \frac{lim_{h→0}[sinx cos x (cos2h – sin2h – 1) + sinh cosh (cos2x – sin2x)] }{h}\)

By referring to Cos2A formula,

\(Cos2A=\cos ^2A-\sin ^2A\)

\(\frac{lim_{h→0}[sinx cos x (cos2h – 1)] / h + limh→0 [sinh cosh cos2x]}{h}\)

\(lim_{h→0}\frac{[sinx cos x (cos2h – 1)]}{h} + lim_{h→0}\frac{[sinh cosh cos2x]}{h}\)

\(2\sin x\cos x\ lim_{h→0}\frac{(\cos2h-1)}{2h}+\cos2x\ lim_{h→0}\frac{\sinh}{h}\times lim_{h→0}\cos\ h\)

= 2 sinx cosx x (0) + cos 2x

= cos 2x

Hence

\(\frac{d}{dx}(\sin x\cos x) =\cos 2x\)

Derivative of Sinx Cosx by Product Rule

We will be proving the result \(\frac{d}{dx}(\sin x\cos x) =\cos 2x\) by Product rule

For the function sinx cosx

f(x) = sinx.

And g(x) = cosx.

\(\frac{d}{dx}(\sin x\cos x)=(\sin x)^{\prime}\cos x+\sin x(\cos x)^{\prime}\)

= cosx. cosx – sinx (-sinx).

As, \(\frac{d}{dx}(\sin x)=\cos x\) and

\(\frac{d}{dx}(\cos x)=\sin x\)

= \(\cos^2x-\sin ^2x\)

= cos 2x.

Hence we obtain the final proof as,

\(\frac{d}{dx}(\sin x\cos x) =\cos 2x\)

Learn about Cos a Cos b

What is the derivative of (sin x) ^(cos x)?

Finding the derivative of \((sin x)^(cos x)\) is a bit more complex, as it involves the chain rule and logarithmic differentiation as well as the product rule. Here's how to do it:

Define the function: Let y = \((sin x)^(cos x)\).

Logarithmic differentiation: We'll first take the natural log of both sides of the equation. This is helpful because it allows us to bring the exponent (cos x) down in front.

ln(y) = cos(x) * ln(sin x)

Differentiate implicitly with respect to x: Now, we'll differentiate each side with respect to x. On the left side, the derivative of ln(y) with respect to y (using the chain rule) is 1/y times dy/dx (which we're trying to find). On the right side, we'll have to use the product rule, where u = cos x and v = ln(sin x). Hence, u' = -sin x and v' = cot x (since the derivative of ln(sin x) is 1/sin x = cot x).

So, (1/y) * dy/dx = [-sin(x) * ln(sin x)] + [cos(x) * cot(x)]

Multiply through by y to isolate dy/dx: Remember that y = \((sin x)^(cos x)\), so:

dy/dx = [-sin(x) * ln(sin x) * \((sin x)^(cos x)\)] + [cos(x) * cot(x) * \((sin x)^(cos x)\)]

Important Notes on Derivative of sin(x)·cos(x)

When you take the derivative of the product of two functions like sin(x) and cos(x), you apply a rule called the product rule. The product rule helps you find the rate of change of two multiplied functions.

In this case, the function is: f(x) = sin(x) · cos(x)

Using the product rule:
Let u = sin(x) and v = cos(x).
Then the derivative is:
f '(x) = u′·v + u·v′ = cos(x)·cos(x) + sin(x)(–sin(x))
= cos²(x) – sin²(x)

This is a well-known identity and can be written as: cos(2x)

So, the derivative of sin(x)·cos(x) is:cos(2x)

You can also understand or derive this using the first principle of derivatives, which uses the limit definition.
Another way is by using trigonometric identities, such as:
2sin(x)cos(x) = sin(2x), so
(1/2)sin(2x) can help simplify certain forms.

Derivative of Sinx Cosx Examples

Example 1: What is the derivative of sinx cosx?

Solution: 

To find the derivative of a function that is the product of two other functions (in this case, sin(x) and cos(x)), you use the product rule. The product rule is defined as follows:

If you have two functions, u(x) and v(x), then the derivative of their product is given by:

(uv)' = u'v + uv'

Now, let's use this rule to find the derivative of sin(x)cos(x).

Define your two functions:

Let u = sin(x) and v = cos(x).

Find the derivatives of these functions:

The derivative of sin(x) is cos(x), so u' = cos(x).

The derivative of cos(x) is -sin(x), so v' = -sin(x).

Substitute these into the product rule:

The derivative of sin(x)cos(x) will be u'v + uv', so:

(sin(x)cos(x))' = cos(x)cos(x) + sin(x)(-sin(x)) = \(cos^2(x)-sin^2(x)\)= cos(2x)

Example 2: What is the derivative of sin(x) + cos(x)?

Solution:

Given:

f(x) = sin(x) + cos(x)

Find:

f'(x) = d/dx [sin(x) + cos(x)]

Solution:

f'(x) = d/dx [sin(x)] + d/dx [cos(x)]

f'(x) = cos(x) - sin(x)

Example 3: Find the derivative of the function x·log(x) using the product rule.

Solution:
Let f(x) = x·log(x)
Here,
u(x) = x and v(x) = log(x)

Now find the derivatives:
u′(x) = 1
v′(x) = 1/x

Apply the product rule:
f′(x) = u′(x)·v(x) + u(x)·v′(x)
= 1·log(x) + x·(1/x)
= log(x) + 1

The derivative of x·log(x) is log(x) + 1

Example 4: Find the derivative of x·sin(x).

Solution:
Let f(x) = x·sin(x)
u(x) = x and v(x) = sin(x)

Now find the derivatives:
u′(x) = 1
v′(x) = cos(x)

Apply the product rule:
f′(x) = u′(x)·v(x) + u(x)·v′(x)
= 1·sin(x) + x·cos(x)
= sin(x) + x·cos(x)

The derivative of x·sin(x) is sin(x) + x·cos(x)

Example 5: Find the derivative of (1 - x)·cos(x).

Solution:
Let f(x) = (1 - x)·cos(x)
Here,
u(x) = (1 - x) and v(x) = cos(x)

Now find the derivatives:
u′(x) = -1
v′(x) = -sin(x)

Apply the product rule:
f′(x) = u′(x)·v(x) + u(x)·v′(x)
= (-1)·cos(x) + (1 - x)(-sin(x))
= -cos(x) - (1 - x)·sin(x)
= -cos(x) - sin(x) + x·sin(x)

The derivative of (1 - x)·cos(x) is -cos(x) - sin(x) + x·sin(x)

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