Find the Derivative - d/dx y=cos( square root of sin(tan(3x)))

Problem

d()/d(x)*cos(√(,sin(tan(3*x))))

Solution

1. Identify the outer function and apply the chain rule for cos(u) where u=√(,sin(tan(3*x)))

d(cos(√(,sin(tan(3*x)))))/d(x)=−sin(√(,sin(tan(3*x))))⋅d(√(,sin(tan(3*x))))/d(x)

2. Apply the chain rule to the square root function, √(,v)=v(1/2) where v=sin(tan(3*x))

d(√(,sin(tan(3*x))))/d(x)=1/(2√(,sin(tan(3*x))))⋅d(sin(tan(3*x)))/d(x)

3. Differentiate the sine function using the chain rule, where the inner function is tan(3*x)

d(sin(tan(3*x)))/d(x)=cos(tan(3*x))⋅d(tan(3*x))/d(x)

4. Differentiate the tangent function and the innermost linear term 3*x

d(tan(3*x))/d(x)=sec2(3*x)⋅(d(3)*x)/d(x)

(d(3)*x)/d(x)=3

5. Combine all the components obtained from the chain rule steps into a single expression.

d(y)/d(x)=−sin(√(,sin(tan(3*x))))⋅1/(2√(,sin(tan(3*x))))⋅cos(tan(3*x))⋅sec2(3*x)⋅3

6. Simplify the expression by grouping the constants and trigonometric terms.

d(y)/d(x)=−(3*sin(√(,sin(tan(3*x))))*cos(tan(3*x))*sec2(3*x))/(2√(,sin(tan(3*x))))

Final Answer

d(cos(√(,sin(tan(3*x)))))/d(x)=−(3*sin(√(,sin(tan(3*x))))*cos(tan(3*x))*sec2(3*x))/(2√(,sin(tan(3*x))))

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