how do i differentiate √secx+tanx ??
I could never remember those formulas, so
f= (cosx)^-1/2 + (sinx)(cosx)^-1
f'= .5 sinx*(cosx)^-3/2 + cosx*(cosx)^(-1)+1*sinx(sinx)cosx)^-2
and that can be simplified. check my work.
or
y = (secx)^1/2 + tanx
dy/dx = (1/2)(secx)^(-1/2)((secx)tanx) + sec^2 x
= (-1/4)(tanx)(secx)^1/2 + sec^2 x
To differentiate the expression √sec(x) + tan(x), we can use the rules of differentiation.
Step 1: Rewrite the expression using trigonometric identities.
Recall that sec(x) is equal to 1/cos(x). Therefore, we can rewrite √sec(x) as √(1/cos(x)). Similarly, tan(x) can be written as sin(x)/cos(x).
Step 2: Simplify the expression.
We can simplify √(1/cos(x)) as 1/√cos(x), and sin(x)/cos(x) can be written as tan(x).
So now, our expression becomes 1/√cos(x) + tan(x).
Step 3: Differentiate the expression term by term.
To differentiate 1/√cos(x), we need to use the chain rule. Let's assign a variable, u, to represent cos(x).
Let u = cos(x).
Now, we rewrite the expression as 1/√u.
To differentiate 1/√u, we can use the power rule. The power rule states that if we differentiate u^n, where n is a constant, the result is nu^(n-1). In our case, n is -1/2.
Differentiating 1/√u, we get (-1/2)u^(-1/2-1) = (-1/2)u^(-3/2), which is -1/(2√u).
Now, we substitute our variable back: -1/(2√cos(x)).
Next, let's differentiate tan(x). The derivative of tan(x) is sec^2(x) using the trigonometric derivative rules.
So, the derivative of 1/√cos(x) is -1/(2√cos(x)), and the derivative of tan(x) is sec^2(x).
Step 4: Combine the derivatives.
Putting it all together, the derivative of √sec(x) + tan(x) is -1/(2√cos(x)) + sec^2(x).
Therefore, the differentiated expression is -1/(2√cos(x)) + sec^2(x).