What is the solution for f^'(x)=-sin(x^5)-78x ?

The solution for f^'(x)=-sin(x^5)-78x is f(x)=integral of-sin(x^5)-78x

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f^'(x)=-sin(x^5)-78x

f^{^{'}} (x) = - sin (x^{5}) - 78 x

f (x) = \int - sin (x^{5}) - 78 x d x

Solution steps

f^{'} (x) = - sin (x^{5}) - 78 x

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

f (x) = \int - sin (x^{5}) - 78 x d x

f (x) = \int - sin (x^{5}) - 78 x d x

y^{^{'}} = 2, x y^{2} y = \frac{1}{( 4 - x ^{2} )}

f^{^{'}} (x) = x^{3} + 8 x - 5 x + 6

y^{^{'}} = \frac{( 2 )}{( x ^{3} ) + 5 x ^{\frac{( 2 )}{( 3 )}}}

x^{2} y^{^{'}} = 1

f^{^{'}} (x) = 9 x^{2} + 10 x

What is the solution for f^'(x)=-sin(x^5)-78x ?
The solution for f^'(x)=-sin(x^5)-78x is f(x)=integral of-sin(x^5)-78x