How do you find the derivative of $f (x) = e^{8 x} + cos (x) sin (x)$ $f(x)=e^(8x)+cos(x)sin(x)$ ?

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Answer 1 · 2016-05-28T08:44:17

$\frac{d}{d x} f (x) = 8 \cdot e^{7 x} \cdot l n e + 1 - 2 {sin}^{2} x$ $d /(d x) f(x)=8*e^(7x)*l n e+1-2sin^2 x$

$f (x) = e^{8 x} + cos (x) sin (x)$ $f(x)=e^(8x)+cos(x)sin(x)$

$\frac{d}{d x} f (x) = 8 \cdot e^{7 x} \cdot l n e - sin x \cdot sin x + cos x \cdot cos x$ $d /(d x) f(x)=8*e^(7x)*l n e-sin x*sin x +cos x *cos x$

$\frac{d}{d x} f (x) = 8 \cdot e^{7 x} \cdot l n e - {sin}^{2} x + {cos}^{2} x$ $d /(d x) f(x)=8*e^(7x)*l n e-sin^2 x+cos^2 x$

${cos}^{2} x = 1 - {sin}^{2} x$ $cos^2 x=1-sin^2 x$

$\frac{d}{d x} f (x) = 8 \cdot e^{7 x} \cdot l n e - {sin}^{2} x + (1 - {sin}^{2} x)$ $d /(d x) f(x)=8*e^(7x)*l n e-sin^2 x+(1-sin^2 x)$

$\frac{d}{d x} f (x) = 8 \cdot e^{7 x} \cdot l n e - {sin}^{2} x + 1 - {sin}^{2} x$ $d /(d x) f(x)=8*e^(7x)*l n e-sin^2 x+1-sin^2 x$

$\frac{d}{d x} f (x) = 8 \cdot e^{7 x} \cdot l n e + 1 - 2 {sin}^{2} x$ $d /(d x) f(x)=8*e^(7x)*l n e+1-2sin^2 x$

How do you find the derivative of f(x)=e8x+cos(x)sin(x) f(x)=e^(8x)+cos(x)sin(x)?