# Differentiate `sin^2x-tanx`

### 2 Answers | Add Yours

The expression `sin^2x - tan x` has to be differentiated.

Using the chain rule, the derivative is:

`2*sin x*cos x - sec^2x`

=> `sin 2x - sec^2x`

**The derivative of sin^2x - tan x is sin 2x - sec^2x**

Let me try to explain the simplest way to find the derivative of the given function.

we have given the function as sin²x - tanx

what about finding the derivative of the function seperately here.

d(sin²x)/dx - d(tanx)/dx

Now, we know that the derivative of the tanx is sec²x. we only need to find the derivative of sin²x.

We will apply the chain rule here.

as we know chain rule states that inner function derivative * outer function derivative. we have inner function as sin²x and outer function as sinx. i.e.

d(sin²x)/dx = d(sin²x)/du * d(sinx)/dx

we know that we can find the derivative of the variable function only with respect to the same variable.i.e for finding derivative of y^2, we need to derivate it with respect to **y**. d(y^2)/**dy** = 2y.

So, since we have d(sin²x)/du. so we need to convert the function sin²x in terms of u to get the derivative.

Let u = sinx then u^2 = sin²x

now, d(sin²x)/du = d(u^2)/du = 2u

We plugged the value u = sinx, so, reverse plugging the value of u = sinx will give.

d(sin²x)/du = d(u^2)/du = 2u = 2sinx

Now, d(sin²x)/dx = d(sin²x)/du * d(sinx)/dx

= 2sinx * d(sinx)/dx

= 2sinx*cosx

So, the complete derivative of sin²x - tanx will be

d(sin²x)/dx - d(tanx)/dx = 2sinx*cosx - sec²x

Hope this will help you!!