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26.   Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

          \frac{4x + 5 \sin x }{3x+ 7 \cos x }

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Given,

f(x)=\frac{4x + 5 \sin x }{3x+ 7 \cos x }

Now, As we know the derivative of any function  

\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}

Also the property 

\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}

Applying those properties,we get

\frac{d(\frac{4x+5\sin x}{3x+7\cos x})}{dx}=\frac{(3x+7\cos x)d(\frac{d(4x+5\sin x)}{dx})-(4x+5\sin x)(\frac{d(3x+7\cos x)}{dx})}{(3x+7\cos x)^2}

\frac{d(\frac{4x+5\sin x}{3x+7\cos x})}{dx}=\frac{(3x+7\cos x)(4+5\cos x)-(4x+5\sin x)(3-7\sin x)}{(3x+7\cos x)^2}

=\frac{12x+28\cos x+15x\cos x+35\cos^2x-12x-15\sin x+28x\sin x+35\sin^2 x}{(3x+7\cos x)^2}

=\frac{12x+28\cos x+15x\cos x-12x-15\sin x+28x\sin x+35(\sin^2 x+\cos^2x)}{(3x+7\cos x)^2}

=\frac{28\cos x+15x\cos x-15\sin x+28x\sin x+35}{(3x+7\cos x)^2}

Posted by

Pankaj Sanodiya

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