## How to calculation the derivative that sin^2x

*Note the in this post we will certainly be spring at separating sin2(x) which is not the exact same as distinguishing sin(2x). Below is our write-up dealing with how to identify sin(2x).You are watching: Sin(x^2) derivative*

There space two techniques that have the right to be used for calculating the derivative the sin^2x.

The very first method is by making use of the product ascendancy for derivatives (since sin2(x) deserve to be written as sin(x).sin(x)).

The second method is by utilizing the chain ascendancy for differentiation.

### Finding the derivative of sin^2x using the product rule

The product rule for differentiation says that the derivative the f(x).g(x) is f’(x)g(x) + f(x).g’(x)

**The Product Rule:****For 2 differentiable functions f(x) and g(x)**

**If F(x) = f(x).g(x)**

**Then the derivative that F(x) is F"(x) = f’(x)g(x) + f(x)g"(x)**

First, allow F(x) = sin2(x)

Then remember the sin2(x) is same to sin(x).sin(x)

So F(x) = sin(x)sin(x)

By setup f(x) and also g(x) as sin(x) way that F(x) = f(x).g(x) and we can apply the product preeminence to discover F"(x)

F"(x) | = f"(x)g(x) + f(x)g"(x) | Product ascendancy Definition |

= f"(x)sin(x) + sin(x)g"(x) | f(x) = g(x) = sin(x) | |

= cos(x)sin(x) + sin(x)cos(x) | f"(x) = g(‘x) = cos(x) | |

= 2sin(x)cos(x) |

Using the product rule,**the derivative that sin^2x is 2sin(x)**cos(x)

### Finding the derivative that sin^2x using the chain rule

The chain rule is helpful for detect the derivative that a duty which could have been identified had it been in x, but it is in the type of one more expression i m sorry could additionally be differentiated if it stood top top its own.

In this case:

We know exactly how to differentiate sin(x) (the prize is cos(x))We know exactly how to identify x2 (the answer is 2x)This means the chain preeminence will enable us to carry out the differentiation of the expression sin^2x.

Using the chain rule to find the derivative the sin^2xAlthough the expression sin2x consists of no parenthesis, we deserve to still see it as a composite role (a function of a function).

We can write sin2x as (sin(x))2.

Now the function is in the form of x2, except it does not have x together the base, rather it has actually another function of x (sin(x)) together the base.

Let’s contact the role of the basic g(x), which means:

g(x) = sin(x)

From this it adheres to that:

sin(x)2 = g(x)2

So if the function f(x) = x2 and the duty g(x) = sin(x), climate the function (sin(x))2 deserve to be created as a composite function.

f(x) = x2f(g(x)) = g(x)2 (but g(x) = sin(x))f(g(x)) = (sin(x))2

Let’s specify this composite duty as F(x):

F(x) = f(g(x)) = (sin(x))2

We can find the derivative the sin^2x (F"(x)) through making use of the chain rule.

**The Chain Rule:****For 2 differentiable functions f(x) and also g(x)**

**If F(x) = f(g(x))**

**Then the derivative the F(x) is F"(x) = f’(g(x)).g’(x)**

Now we have the right to just plug f(x) and also g(x) right into the chain rule.

**How to uncover the derivative that sin^2x using the Chain Rule:**

F"(x) | = f"(g(x)).g"(x) | Chain dominance Definition |

= f"(g(x))(cos(x)) | g(x) = sin(x) ⇒ g"(x) = cos(x) | |

= (2sin(x)).(cos(x)) | f(g(x)) = (sin(x))2 ⇒ f"(g(x)) = 2sin(x) | |

= 2sin(x)cos(x) |

Using the chain rule,**the derivative the sin^2x is 2sin(x)cos(x)**

*(Note – utilizing the trigonometric identification 2cos(x)sin(x) = sin(2x), the derivative the sin^2x can also be created as sin(2x))*

Finally, just a keep in mind on syntax and also notation: sin^2x is periodically written in the forms below (with the derivative as per the calculations above). Simply be aware that not every one of the forms below are mathematically correct.

sin2x | ► Derivative of sin2x = 2sin(x)cos(x) |

sin^2(x) | ► Derivative that sin^2(x) = 2sin(x)cos(x) |

sin 2 x | ► Derivative of sin 2 x = 2sin(x)cos(x) |

(sinx)^2 | ► Derivative that (sinx)^2 = 2sin(x)cos(x) |

sin squared x | ► Derivative of sin squared x = 2sin(x)cos(x) |

sinx2 | ► Derivative the sinx2 = 2sin(x)cos(x) |

sin^2 | ► Derivative of sin^2 = 2sin(x)cos(x) |

## The 2nd Derivative the sin^2x

To calculate the 2nd derivative that a function, friend just distinguish the first derivative.

From above, we discovered that the very first derivative of sin^2x = 2sin(x)cos(x). So to find the 2nd derivative the sin^2x, we just need to differentiate 2sin(x)cos(x)

We have the right to use the product preeminence to discover the derivative of 2sin(x)cos(x).

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We can set f(x) = 2sin(x) and also g(x) = cos(x) and also apply the product rule to discover the derivative the f(x).g(x) = 2sin(x)cos(x)

The product rules claims that the derivative the f(x).g(x) is equal to f’(x)g(x) + f(x)g’(x).

2cos(x)cos(x) + 2sin(x)(-sin(x))= 2cos2(x) – 2sin2(x)= 2(cos2(x) – sin2(x))

Using the trigonometric dual angle identification cos(2x) = cos2(x) – sin2(x), we can rewrite this as

= 2cos(2x)

► **The second derivative of sin^2x is 2cos(2x)**

*Interestingly, the second derivative that sin2x is same to the first derivative of sin(2x). *