Derivative of tan(x2+π4)\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}

The calculator will find the derivative of tan(x2+π4)\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}, with steps shown.

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Your Input

Find ddx(tan(x2+π4))\frac{d}{dx} \left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right).

Solution

The function tan(x2+π4)\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)} is the composition f(g(x))f{\left(g{\left(x \right)} \right)} of two functions f(u)=tan(u)f{\left(u \right)} = \tan{\left(u \right)} and g(x)=x2+π4g{\left(x \right)} = \frac{x}{2} + \frac{\pi}{4}.

Apply the chain rule ddx(f(g(x)))=ddu(f(u))ddx(g(x))\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right):

(ddx(tan(x2+π4)))=(ddu(tan(u))ddx(x2+π4)){\color{red}\left(\frac{d}{dx} \left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\tan{\left(u \right)}\right) \frac{d}{dx} \left(\frac{x}{2} + \frac{\pi}{4}\right)\right)}

The derivative of the tangent is ddu(tan(u))=sec2(u)\frac{d}{du} \left(\tan{\left(u \right)}\right) = \sec^{2}{\left(u \right)}:

(ddu(tan(u)))ddx(x2+π4)=(sec2(u))ddx(x2+π4){\color{red}\left(\frac{d}{du} \left(\tan{\left(u \right)}\right)\right)} \frac{d}{dx} \left(\frac{x}{2} + \frac{\pi}{4}\right) = {\color{red}\left(\sec^{2}{\left(u \right)}\right)} \frac{d}{dx} \left(\frac{x}{2} + \frac{\pi}{4}\right)

Return to the old variable:

sec2((u))ddx(x2+π4)=sec2((x2+π4))ddx(x2+π4)\sec^{2}{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(\frac{x}{2} + \frac{\pi}{4}\right) = \sec^{2}{\left({\color{red}\left(\frac{x}{2} + \frac{\pi}{4}\right)} \right)} \frac{d}{dx} \left(\frac{x}{2} + \frac{\pi}{4}\right)

The derivative of a sum/difference is the sum/difference of derivatives:

sec2(x2+π4)(ddx(x2+π4))=sec2(x2+π4)(ddx(x2)+ddx(π4))\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{dx} \left(\frac{x}{2} + \frac{\pi}{4}\right)\right)} = \sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{dx} \left(\frac{x}{2}\right) + \frac{d}{dx} \left(\frac{\pi}{4}\right)\right)}

The derivative of a constant is 00:

((ddx(π4))+ddx(x2))sec2(x2+π4)=((0)+ddx(x2))sec2(x2+π4)\left({\color{red}\left(\frac{d}{dx} \left(\frac{\pi}{4}\right)\right)} + \frac{d}{dx} \left(\frac{x}{2}\right)\right) \sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} = \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(\frac{x}{2}\right)\right) \sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}

Apply the constant multiple rule ddx(cf(x))=cddx(f(x))\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right) with c=12c = \frac{1}{2} and f(x)=xf{\left(x \right)} = x:

sec2(x2+π4)(ddx(x2))=sec2(x2+π4)(ddx(x)2)\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{dx} \left(\frac{x}{2}\right)\right)} = \sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{\frac{d}{dx} \left(x\right)}{2}\right)}

Apply the power rule ddx(xn)=nxn1\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1} with n=1n = 1, in other words, ddx(x)=1\frac{d}{dx} \left(x\right) = 1:

sec2(x2+π4)(ddx(x))2=sec2(x2+π4)(1)2\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{2} = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} {\color{red}\left(1\right)}}{2}

Simplify:

sec2(x2+π4)2=11sin(x)\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} = \frac{1}{1 - \sin{\left(x \right)}}

Thus, ddx(tan(x2+π4))=11sin(x)\frac{d}{dx} \left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right) = \frac{1}{1 - \sin{\left(x \right)}}.

Answer

ddx(tan(x2+π4))=11sin(x)\frac{d}{dx} \left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right) = \frac{1}{1 - \sin{\left(x \right)}}A