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Popular >

derivative of y=f(x)=tan(sqrt(x)+1)

Solution

derivative of

y = f (x) = tan (x + 1)

Solution

\frac{sec ^{2} ( x + 1 )}{2 x}

Solution steps

\frac{d}{d x} (tan (x + 1))

Apply the chain rule:

sec^{2} (x + 1) \frac{d}{d x} (x + 1)

= sec^{2} (x + 1) \frac{d}{d x} (x + 1)

\frac{d}{d x} (x + 1) = \frac{1}{2 x}

= sec^{2} (x + 1) \frac{1}{2 x}

Simplify

sec^{2} (x + 1) \frac{1}{2 x} : \frac{sec ^{2} ( x + 1 )}{2 x}

= \frac{sec ^{2} ( x + 1 )}{2 x}

Popular Examples

tangent of f(x)=7x-6sqrt(x),\at x=1 tangent of

f (x) = 7 x - 6 x, at x = 1

limit as x approaches 10 of 12x

x \to 10 lim (12 x)

3x^2ydx+(y+x^3)dy=0

3 x^{2} y d x + (y + x^{3}) d y = 0

derivative of ln(1/(1+e^{-x}))

\frac{d}{d x} (ln (\frac{1}{1 + e ^{- x}}))

derivative of (sqrt(x)/(1+x^{3/2)})

\frac{d}{d x} (\frac{x}{1 + x ^{\frac{3}{2}}})

Frequently Asked Questions (FAQ)

What is the derivative of y=f(x)=tan(sqrt(x)+1) ?
The derivative of y=f(x)=tan(sqrt(x)+1) is (sec^2(sqrt(x)+1))/(2sqrt(x))
What is the first derivative of y=f(x)=tan(sqrt(x)+1) ?
The first derivative of y=f(x)=tan(sqrt(x)+1) is (sec^2(sqrt(x)+1))/(2sqrt(x))